Proving whether or not this limit exists

That limit exists and it is equal to $0$, since $\lvert x\rvert,\lvert y\rvert\leqslant\sqrt{x^2+y^2}$ and therefore$$\left\lvert\frac{3x^2y}{x^2+y^2}\right\rvert\leqslant3\frac{\sqrt{x^2+y^2}^3}{x^2+y^2}=3\sqrt{x^2+y^2}.$$So, given $\varepsilon>0$, take $\delta=\frac\varepsilon3$ and then$$\bigl\lVert(x,y)\bigr\rVert<\delta\implies\left\lvert\frac{3x^2y}{x^2+y^2}\right\rvert<\varepsilon.$$