Evaluating $\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}$ 
$$\lim_{(x,y)\to(0,0)}\frac{x^2+y^2}{\sin^2y+\ln(1+x^2)}$$

If I use a specific path I know I can use Cauchy Theorem to get a number, but how do I prove this for all paths? Thank you!

You can always pass to polar coordinates to compute 2d limits:
Set $x=r \cos\theta , y=r\sin \theta$, and now we need to compute
$$\lim_{r\to 0} \dfrac{r^2}{\sin^2(r\sin \theta) + \ln (1+r^2\cos^2\theta)}$$ 
Since this is a limit for the variable $r$, you can use L'Hôpital taking derivatives w.r.t. $r$. Note that if we end with something depending on $\theta$, the limit does not exist, becasuse the limit (if exists) is unique.
After L'Hôpital and some basic manipulation, we end with
$$\lim_{r\to 0} \dfrac{r^2}{\sin^2(r\sin \theta) + \ln (1+r^2\cos^2\theta)} = \lim_{r\to 0} \dfrac{2r(1+r^2\cos^2\theta)}{\sin (2r\sin\theta) \sin \theta(1+r^2\cos^2\theta) + 2r\cos^2\theta}$$
You use the always useful $\sin (ar) \approx ar$ when $r \approx 0$ to rewrite the denominator of the RHS:
$$\lim_{r\to 0} \dfrac{2r(1+r^2\cos^2\theta)}{(2r\sin\theta)\sin\theta(1+r^2\cos^2\theta) + 2r\cos^2\theta} = \lim_{r\to 0} \dfrac{2r(1+r^2\cos^2\theta)}{2r\sin^2 \theta(1+r^2\cos^2\theta) + 2r\cos^2\theta}= \dfrac{1}{\sin^2 \theta + \cos^2\theta} = 1$$
