It is clear that $f(0,0)$ does not exist, and therefore the function is discontinuous. Now to know if the discontinuity is avoidable or inevitable, we must see that the limit at that point exists or not. To see the above we will use polars.
\begin{align*}
\lim_{(x,y)\rightarrow(0,0)}\frac{5xy^2}{x^2+y^2}&=\lim_{r\rightarrow0}\frac{5(r\cos(\theta))(r\sin(\theta))^2}{(r\cos(\theta))^2+(r\sin(\theta))^2}\\
& =\lim_{r\rightarrow0}\frac{5r^3\cos(\theta)\sin^2(\theta)}{r^2(\cos^2(\theta)+\sin^2(\theta))}\\
& =\lim_{r\rightarrow0}\frac{5r^3\cos(\theta)\sin^2(\theta)}{r^2}\\
& =\lim_{r\rightarrow0}5r\cos(\theta)\sin^2(\theta)=0.
\end{align*}
Now let's test for the definition $\epsilon-\delta$, to see that the limit exists and is zero.
\begin{align*}
\left|\frac{5xy^2}{x^2+y^2}-0\right| & =\left|\frac{5xy^2}{x^2+y^2}\right|\\
& =\frac{5|x|y^2}{x^2+y^2}\\
& \leq 5|x| \text{ since $y^2\leq x^2+y^2$}\\
& \leq 5\sqrt{x^2+y^2} \text{ since $|x|\leq\sqrt{x^2+y^2}$}\\
& <5\delta = \epsilon.
\end{align*}
Therefore we have:
\begin{align*}
\forall\epsilon>0, \exists\delta>0 \text{ such that, if } 0<\sqrt{x^2+y^2}<\delta \Rightarrow |f(x,y)-0| &\leq5 \sqrt{x^2+y^2}\\
&<5 \delta = 5\left(\frac{\epsilon}{5}\right)=\epsilon.
\end{align*}
Thus the limit exists and is zero, therefore the discontinuity of $f$ is avoidable.
I think this is the correct solution, I await your comments. If anyone has a different solution or correction of my work I will be grateful.
Best Answer
Both your proofs (the analytic proof, and the proof using polar coordinates) that $\lim_{(x,y) f(x,y)\rightarrow (0,0)}=0$ are correct.