As I understand the question, you are also interested in how one finds functions for Rouché. That is why I give this answer, even though I cannot prove the last inequality (which is however true).
For Rouché you want a function which is close enough to $f(z) = z^2 e^{z}-z$ for $z=2e^{i\phi}$, but simple enough such that you know how many zeros there are in circle with $|z| \leq 2$. I cannot give you a general recipe. But here is how I proceed to find it:
The first obvious choice $g(z) =z^2 e^{z}$ (for which we know the number of zeros) fails for $z$ around $-2$, simply because the $g(z)$ becomes exponential small whereas in $f(z)$ the $-z$ term dominates. Therefore, you want some other term in $g(z)$ which takes around $-2$ whereas you still need to have an $e^{z}$ in order that the function approximates $f(z)$ around $2$ correctly ($f$ is quite large for $z=2$ and then quickly falls off before at some value the $z$ term takes over).
My second guess therefore is $g(z)= z^2 ( e^{z} + c)$ with some $c>0$. Playing around a bit, I see that the value $c=1/2$ seems to work.
Now what one has to proof: we have to show that $|f(z) - g(z)| < |g(z)|$ for $z=2 e^{i\phi}$. The left hand side is $\left|z(\frac{1}{2}+z)\right| = 4 \left| \cos (\phi/2)\right|$. The right hand side can be simplified to $|g(z)|^2= 16 \left|e^{z} +\frac{1}{2}\right|^2 = 16 \left[\frac{1}{4} + e^{4 \cos \phi} + e^{2 \cos \phi} \cos(2 \sin \phi)\right].$
Taking the difference $ \frac{|g(z)|^2 -|f(z) - g(z)|^2}{16} = \frac{1}{4} + e^{4 \cos \phi} + e^{2 \cos \phi} \cos(2 \sin \phi) - \cos^2 (\phi/2) \geq 0.07 .$ The last estimate was obtained by plotting the function. The minimum is attained around $\phi=1.9$. Too make the argument complete one should prove that the difference is larger than zero (which I couldn't do so far). Then we know that $f(z)$ and $g(z)$ have the same number of zeros with $|z| \leq 2$. As $g(z)$ has only two zeros (two times $z=0$), the same hold for $f(z)$.