Can someone please prove this limit via the squeeze theorem

You have for $(x,y) \neq (0,0)$

$$0 \le \left\vert \frac{5x^2y}{x^2+8y^2}\right\vert \le \left\vert\frac{5x^2y}{x^2+y^2} \right\vert\le \frac{5}{2} \vert x \vert \left\vert\frac{\vert x y \vert}{x^2+y^2} \right\vert \le \frac{5\vert x \vert }{2}$$

as $\vert x y \vert \le \frac{x^2+y^2}{2}$ for all $(x,y) \in \mathbb R$.

As $\lim\limits_{(x,y) \to (0,0)} \vert x \vert = 0$, you get the desired conclusion by the squeeze theorem.

$$\left|\frac{5x^2y}{x^2+8y^2}\right|=5|y|\left|\frac{x^2}{x^2+8y^2}\right|\le 5|y|\to0.$$

We have

$$\left|\frac{5x^2y}{x^2+8y^2}\right|=\frac{5x^2|y|}{x^2+8y^2}\le \frac{5x^2|y|+5|y|y^2}{x^2+y^2}= 5|y|\frac{x^2+y^2}{x^2+y^2}=5|y| \to 0$$

or also more simply

$$\left|\frac{5x^2y}{x^2+8y^2}\right| =5|y|\frac{x^2}{x^2+8y^2} \le5|y| \to 0$$