Relation between Output voltage and current in a transresistance amplifier
You already know \$V_1\$. And given your edited/added approach to solving the problem, which works too, I've no problem adding the follow-up to my earlier suggestion that you use nodal analysis.

So just do the nodal for \$V_1\$:

$$\begin{align*}
\frac{V_1}{R_2}+\frac{V_1}{R_3}&amp;=i_s+\frac{v_o}{R_3}\\\\
V_1\cdot\left(\frac{1}{R_2}+\frac{1}{R_3}\right)&amp;=i_s+\frac{v_o}{R_3}
\end{align*}$$

That's the nodal for \$V_1\$. But you also know that \$V_1=-i_s\cdot R_1\$. (You already said so.) So:

$$\begin{align*}
-i_s\cdot R_1\cdot\left(\frac{1}{R_2}+\frac{1}{R_3}\right)&amp;=i_s+\frac{v_o}{R_3}\\\\
-i_s-i_s\cdot R_1\cdot\left(\frac{1}{R_2}+\frac{1}{R_3}\right)&amp;=\frac{v_o}{R_3}\\\\
-i_s\cdot\left[1+ R_1\cdot\left(\frac{1}{R_2}+\frac{1}{R_3}\right)\right]&amp;=\frac{v_o}{R_3}\\\\
v_o&amp;=-i_s\cdot R_3\cdot\left[1+ R_1\cdot\left(\frac{1}{R_2}+\frac{1}{R_3}\right)\right]\\\\
\frac{v_o}{i_s}&amp;=- R_3\cdot\left[1+ R_1\cdot\left(\frac{1}{R_2}+\frac{1}{R_3}\right)\right]\\\\
\frac{v_o}{i_s}&amp;=- \left(R_3+ \frac{R_1 R_3}{R_2}+R_1\right)\\\\
\frac{v_o}{i_s}&amp;=- R_1\cdot\left(1+\frac{R_3}{R_1}+ \frac{R_3}{R_2}\right)
\end{align*}$$

Which amounts to what you said you needed to prove.

However, it wouldn't hurt to go one more step:

$$\begin{align*}
\frac{v_o}{i_s}&amp;=- R_1\cdot R_3\left(\frac{1}{R_1}+ \frac{1}{R_2}+\frac{1}{R_3}\right)\\\\
&amp;=-\frac{R_1\cdot R_3}{R_1\:\mid\mid\: R_2\:\mid\mid\: R_3}
\end{align*}$$

Since all three resistors are attached to voltage sources, and a common node, you'd expect that they are in some way parallel to each other. The above equation makes that fact explicit.