To prove that Close Loop transfer function is sensitive to perturbation in H(s) but not to perturbations in G(s)
In a comment I said, "First solve for \$\frac{C_s}{R_s}\$. Then take the derivative with respect to the two variables of interest and compare." And this is the direction I'll take below. (Others may have different approaches. But that's mine.)

Again, I'd like to recommend that you use the simultaneous linear equations solution process. (It's good to keep that skill in continual practice so that it just rolls from your fingers, fluently.) But in this case, it's also fairly simple algebra. So I'll do it differently.

Here's your diagram with two labeled nodes (the one in between K and G isn't important.) Obviously, \$C_s\$ is the same as \$N_2\$. But \$R_s\$ isn't the same as \$N_1\$.



So:

$$\begin{align*}
N_1&amp;=R_s-H\cdot N_2\\
N_2&amp;=K\cdot G\cdot N_1
\end{align*}$$

As I wrote above, you could solve that simultaneously and get the answer for \$N_2\$ (and then divide by \$R_s\$ to get the transfer function.) But you've already indicated in your earlier question that simultaneous solutions aren't a strength of yours and that Cramer's Rule is something you vaguely remember and need to refresh yourself about. So, as I also wrote above I'll just substitute and solve.

Let's do that (please accept my apologies for shortening \$H_s\$ to \$H\$ and \$G_s\$ to \$G\$):

$$\begin{align*}
N_2&amp;=K\cdot G\cdot N_1\\
&amp;=K\cdot G\cdot \left(R_s-H\cdot N_2\right)\\
&amp;=K\cdot G\cdot R_s-K\cdot G\cdot H\cdot N_2\\
N_2+K\cdot G\cdot H\cdot N_2&amp;=K\cdot G\cdot R_s\\
N_2\cdot\left(1+K\cdot G\cdot H\right)&amp;=K\cdot G\cdot R_s\\\\
\therefore\quad N_2&amp;=\frac{K\cdot G\cdot R_s}{1+K\cdot G\cdot H}
\end{align*}$$

Now, divide out \$R_s\$ to get the transfer function of:

$$T=\frac{C_s}{R_s}=\frac{K\cdot G}{1+K\cdot G\cdot H}$$

The question states that you have a "feedback control system with high gain K." This is a clue for you that you need to keep in mind. They didn't waste words when they said "high gain." They meant for you to use that fact in your analysis. You need to watch for these kinds of phrases in such questions. They are important and you need to use your mind and consider them.

The question also says that you are supposed to "prove that the closed loop transfer function is sensitive to perturbation in \$H_s\$, but not to perturbations in \$G_s\$." Any time you see the phrase "sensitive to" you need to start immediately thinking in terms of derivatives. (There are several different meanings that this phrase might take on, depending upon the context. Regardless, derivatives of some kind are usually implied.)

If you also are not thoroughly familiar with derivatives, and this should include Implicit Differentiation, the Differential Operator, and the Shift Theorem, then you need to brush up on these. You will need them. However, in this case, we can keep it simple and just look at the partial derivatives of the transfer function with respect to selected variables. Those variables will, of course, be the ones discussed in the problem statement: \$G\$ and \$H\$.

$$\begin{align*}
\frac{\text{d}\,T}{\text{d}\,G}&amp;=\frac{K}{1+K\cdot G\cdot H}-\frac{K^2\cdot G\cdot H}{\left(1+K\cdot G\cdot H\right)^2}\\\\
&amp;=\frac{1}{\frac{1}{K}+G\cdot H}-\frac{G\cdot H}{\left(\frac{1}{K}+G\cdot H\right)^2}\tag{1}\\\\
\frac{\text{d}\,T}{\text{d}\,H}&amp;=-\frac{K^2\cdot G\cdot H}{\left(1+K\cdot G\cdot H\right)^2}\\\\
&amp;=-\frac{G\cdot H}{\left(\frac{1}{K}+G\cdot H\right)^2}\tag{2}
\end{align*}$$

Since \$K\$ is considered large in the above context, we can simplify the above equations:

$$\begin{align*}
\frac{\text{d}\,T}{\text{d}\,G}&amp;=\frac{1}{\frac{1}{K}+G\cdot H}-\frac{G\cdot H}{\left(\frac{1}{K}+G\cdot H\right)^2}\\\\
&amp;=\frac{1}{G\cdot H}-\frac{G\cdot H}{\left(G\cdot H\right)^2}\\\\
&amp;=\frac{1}{G\cdot H}-\frac{1}{G\cdot H}\\\\
&amp;= 0\tag{$1^*$}\\\\
\frac{\text{d}\,T}{\text{d}\,H}&amp;=-\frac{G\cdot H}{\left(\frac{1}{K}+G\cdot H\right)^2}\\\\
&amp;=-\frac{G\cdot H}{\left(G\cdot H\right)^2}\\\\
&amp;=-\frac{1}{G\cdot H}\tag{$2^*$}
\end{align*}$$

Since equation \$\left(1^*\right)\$ is approximately zero, it follows that the transfer function is insensitive to perturbations in \$G_s\$. But since  equation \$\left(2^*\right)\$ depends in part on \$H_s\$, it follows that the transfer function is sensitive to perturbations in \$H_s\$.

(Note: In the above process, I've assumed that the product, \$K\cdot G\$, is positive. If the product is negative, then my development of equation \$\left(1^*\right)\$ as being approximately zero is incorrect and the sign of the simplified equation \$\left(2^*\right)\$ would be wrong. Keep that also in mind.)