How to prepare observation table and plot frequency response of Butterworth Sallen-Key High Pass Filter
First, I will present a method that uses Mathematica to solve this problem. When I was studying this stuff I used the method all the time (without using Mathematica of course).

Well, we are trying to analyze the following opamp-circuit:



simulate this circuit &ndash; Schematic created using CircuitLab
When we use and apply KCL, we can write the following set of equations:
$$\text{I}_3=\text{I}_1+\text{I}_2\tag1$$
When we use and apply Ohm's law, we can write the following set of equations:
$$
\begin{cases}
\text{I}_1=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}\\
\\
\text{I}_2=\frac{\text{V}_3-\text{V}_1}{\text{R}_2}\\
\\
\text{I}_3=\frac{\text{V}_1-\text{V}_2}{\text{R}_3}\\
\\
\text{I}_3=\frac{\text{V}_2}{\text{R}_4}
\end{cases}\tag2
$$
Substitute \$(2)\$ into \$(1)\$, in order to get:
$$
\begin{cases}
\frac{\text{V}_1-\text{V}_2}{\text{R}_3}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_3-\text{V}_1}{\text{R}_2}\\
\\
\frac{\text{V}_2}{\text{R}_4}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_3-\text{V}_1}{\text{R}_2}
\end{cases}\tag3
$$
Now, using an ideal opamp, we know that:
$$\text{V}_x:=\text{V}_+=\text{V}_-=\text{V}_2=\text{V}_3\tag4$$
So we can rewrite equation \$(3)\$ as follows:
$$
\begin{cases}
\frac{\text{V}_1-\text{V}_x}{\text{R}_3}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_x-\text{V}_1}{\text{R}_2}\\
\\
\frac{\text{V}_x}{\text{R}_4}=\frac{\text{V}_\text{i}-\text{V}_1}{\text{R}_1}+\frac{\text{V}_x-\text{V}_1}{\text{R}_2}
\end{cases}\tag5
$$
Now, we can solve for the transfer function:
$$\mathcal{H}:=\frac{\text{V}_x}{\text{V}_\text{i}}=\frac{\text{R}_2\text{R}_4}{\text{R}_1\left(\text{R}_2+\text{R}_3\right)+\text{R}_2\left(\text{R}_3+\text{R}_4\right)}\tag6$$
Where I used the following Mathematica-code:
In[1]:=Clear[&quot;Global`*&quot;];
V2 = Vx;
V3 = Vx;
FullSimplify[
 Solve[{I3 == I1 + I2, I1 == (Vi - V1)/R1, I2 == (V3 - V1)/R2, 
   I3 == (V1 - V2)/R3, I3 == V2/R4}, {I1, I2, I3, V1, Vx}]]

Out[1]={{I1 -&gt; ((R2 + R3) Vi)/(R1 (R2 + R3) + R2 (R3 + R4)), 
  I2 -&gt; -((R3 Vi)/(R1 (R2 + R3) + R2 (R3 + R4))), 
  I3 -&gt; (R2 Vi)/(R1 (R2 + R3) + R2 (R3 + R4)), 
  V1 -&gt; (R2 (R3 + R4) Vi)/(R1 (R2 + R3) + R2 (R3 + R4)), 
  Vx -&gt; (R2 R4 Vi)/(R1 (R2 + R3) + R2 (R3 + R4))}}

My equation was also confirmed using LTspice.

When we want to apply the derivation from above to your circuit we need to use Laplace transform (I will use lower case function names for the functions that are in the (complex) s-domain, so \$\text{y}\left(\text{s}\right)\$ is the Laplace transform of the function \$\text{Y}\left(t\right)\$):

$$\text{R}_1=\frac{1}{\text{sC}_1}\tag7$$
$$\text{R}_3=\frac{1}{\text{sC}_2}\tag8$$

So, we can rewrite the transfer function as:
$$\mathscr{H}\left(\text{s}\right)=\frac{\text{R}_2\text{R}_4}{\frac{1}{\text{sC}_1}\left(\text{R}_2+\frac{1}{\text{sC}_2}\right)+\text{R}_2\left(\frac{1}{\text{sC}_2}+\text{R}_4\right)}=$$
$$\frac{\text{C}_1\text{C}_2\text{R}_2\text{R}_4\text{s}^2}{\text{C}_1\text{C}_2\text{R}_2\text{R}_4\text{s}^2+\text{R}_2\left(\text{C}_1+\text{C}_2\right)\text{s}+1}\tag9$$
Now, when working with sinusoidal signals we can use \$\text{s}:=\text{j}\omega\$ (where \$\text{j}^2=-1\$ and \$\omega=2\pi\text{f}\$ with \$\text{f}\$ is the frequency of the input signal in Hertz). So we get:
$$\underline{\mathscr{H}}\left(\text{j}\omega\right)=\frac{\text{C}_1\text{C}_2\text{R}_2\text{R}_4\left(\text{j}\omega\right)^2}{\text{C}_1\text{C}_2\text{R}_2\text{R}_4\left(\text{j}\omega\right)^2+\text{R}_2\left(\text{C}_1+\text{C}_2\right)\text{j}\omega+1}=$$
$$\frac{\text{C}_1\text{C}_2\text{R}_2\text{R}_4\omega^2}{\text{C}_1\text{C}_2\text{R}_2\text{R}_4\omega^2-1-\text{R}_2\left(\text{C}_1+\text{C}_2\right)\omega\text{j}}\tag{10}$$
Now, you want to find the absolute value:
$$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{\text{C}_1\text{C}_2\text{R}_2\text{R}_4\omega^2}{\sqrt{\left(\text{C}_1\text{C}_2\text{R}_2\text{R}_4\omega^2-1\right)^2+\left(\text{R}_2\left(\text{C}_1+\text{C}_2\right)\omega\right)^2}}\tag{11}$$

Assuming that \$\text{C}_1=\text{C}_2:=\text{C}\$, we can simplify \$(11)\$ as follows:
$$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{\text{C}^2\text{R}_2\text{R}_4\omega^2}{\sqrt{\left(\text{C}^2\text{R}_2\text{R}_4\omega^2-1\right)^2+\left(2\text{C}\text{R}_2\omega\right)^2}}\tag{12}$$
Solving for the maximum gives:
$$\frac{\partial\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|}{\partial\omega}=0\space\Longrightarrow\space\omega=\frac{1}{\text{C}\sqrt{\text{R}_2\left(\text{R}_4-2\text{R}_2\right)}}\tag{13}$$
With the condition that \$2\text{R}_2&lt;\text{R}_4\$.
The code I used to find it is:
In[2]:=Clear[&quot;Global`*&quot;];
C1 = c;
C2 = c;
FullSimplify[
 Solve[{D[(C1*C2*R2*R4*\[Omega]^2)/
      Sqrt[(C1*C2*R2*R4*\[Omega]^2 - 
           1)^2 + (R2*(C1 + C2)*\[Omega])^2], \[Omega]] == 
    0, \[Omega] &gt; 0 &amp;&amp; c &gt; 0 &amp;&amp; R2 &gt; 0 &amp;&amp; R4 &gt; 0}, \[Omega]], 
 Assumptions -&gt; c &gt; 0 &amp;&amp; R2 &gt; 0 &amp;&amp; R4 &gt; 0]

Out[2]={{\[Omega] -&gt; 
   ConditionalExpression[1/(c Sqrt[R2 (-2 R2 + R4)]), 2 R2 &lt; R4]}}

Because we want one cut-off frequency, we must have that \$2\text{R}_2\ge\text{R}_4\$. Now, I used Mathematica to find that the following component values work:
$$\text{C}=\frac{\sqrt{\frac{1}{10}\left(\sqrt{3996002}+1999\right)}}{200000000 \pi }\approx3.1823\cdot10^{-8}\space\text{F}\space\wedge\space$$
$$\text{R}_2=1000000\space\Omega\space\wedge\space\text{R}_4=1000\space\Omega\tag{14}$$
Code:
In[3]:=Clear[&quot;Global`*&quot;];
C1 = c;
C2 = c;
FullSimplify[
 Solve[{((C1*C2*R2*R4*(10*1000*2*Pi)^2)/
      Sqrt[(C1*C2*R2*R4*(10*1000*2*Pi)^2 - 
           1)^2 + (R2*(C1 + C2)*(10*1000*2*Pi))^2]) == 1/Sqrt[2], 
   10^(-9) &lt; c &lt;= 10^(-3) &amp;&amp; 10^3 &lt;= R2 &lt;= 10^6 &amp;&amp; 
    10^3 &lt;= R4 &lt;= 10^6 &amp;&amp; 2 R2 &gt;= R4}, {c, R2, R4}]]

Out[3]={{R4 -&gt; ConditionalExpression[(-1 + Sqrt[
     2 + 1600000000 c^2 \[Pi]^2 R2^2])/(400000000 c^2 \[Pi]^2 R2), 
    c &gt; 0 &amp;&amp; 
     R2 &lt;= 1000000 &amp;&amp; ((20000000 c \[Pi] &gt; Sqrt[1 + Sqrt[2]] &amp;&amp; 
         2000000000 c \[Pi] &lt; Sqrt[10 (1999 + Sqrt[3996002])] &amp;&amp; 
         1/(400000000000 c^2 \[Pi]^2 - 
           40000 c \[Pi] Sqrt[-1 + 200000000000000 c^2 \[Pi]^2]) &lt;= 
          R2) || (40000000 c \[Pi] &gt; Sqrt[2] &amp;&amp; R2 &gt;= 1000 &amp;&amp; 
         20000000 c \[Pi] &lt;= Sqrt[1 + Sqrt[2]]) || (1000000000 c &gt; 1 &amp;&amp;
          40000 c \[Pi] R2 &gt;= Sqrt[2] &amp;&amp; 
         40000000 c \[Pi] &lt;= Sqrt[2]))]}, {c -&gt; Sqrt[
   1/10 (1999 + Sqrt[3996002])]/(200000000 \[Pi]), R2 -&gt; 1000000, 
  R4 -&gt; 1000}}