I have a complex integral to solve where we want to integrate $f(z) = {z\over(z-1)(z^2+1)}$ over the contour $C=16x^2+y^2=4$
My problem is that I don't know how to draw the contour circle and find $R$ to find out whether my poles are inside the region or not.
Well, we have $$f(z) = {z\over(z-1)(z^2+1)} = {z\over(z-1)(z-i)(z+i)}$$ This has poles at $z=1, \pm i$. Your contour is an ellipse given by $\left(4x\right)^{2}+y^{2}=4$, which is a circle with its sides (left right) squashed by a scale factor of $4$, making it land at $1/2$.
From this, we can see that the only poles inside the contour are $z=\pm i$. Here is why. Your three poles lie on the real and imaginary axis. Your ellipse is a circle squashed about the real axis. This means, the imaginary endpoints end at $\pm 2i$, and the real axis endpoints are $\pm \frac12$. This makes it obvious that the poles on the imaginary axis $\pm i$ lie inside the contour, and the real pole lies outside the contour since $\frac12 < 1$.
Now use the residue theorem
$$\text{your integral over your contour}= 2\pi i \operatorname{Res}(f(z), z=\pm i)$$
to calculate the value.