Answer
$\lim_{(x,y)\to (0,0) }f(x,y)=0$
$\lim_{(x,y)\to (0,0) }g(x,y)$ does not exist
Work Step by Step
Let $x=r\cos \theta $, $y=r\sin \theta $, then
\begin{align*}
\lim_{(x,y)\to (0,0) }f(x,y)&=\lim_{(x,y)\to (0,0) }\frac{x^3}{x^2+y^2}\\
&=\lim_{r\to 0} \frac{r^3\cos^3\theta}{r^2\cos^2\theta+ r^2\sin^2\theta}\\
&= \lim_{r\to 0} \frac{r^3\cos^3\theta}{r^2 }\\
&= \lim_{r\to 0} r \cos^3\theta \\
&=0
\end{align*}
and
\begin{align*}
\lim_{(x,y)\to (0,0) }g(x,y)&=\lim_{(x,y)\to (0,0) }\frac{x^2}{x^2+y^2}\\
&=\lim_{r\to 0} \frac{r^2\cos^2\theta}{r^2\cos^2\theta+ r^2\sin^2\theta}\\
&= \lim_{r\to 0} \frac{r^2\cos^2\theta}{r^2 }\\
&= \lim_{r\to 0} \cos^2\theta \\
&=0
\end{align*}
Since $\lim_{(x,y)\to (0,0) }g(x,y) $ depends on theta, then $\lim_{(x,y)\to (0,0) }g(x,y)$ does not exist.
