Let $f(x)=x^2$. Since $f$ is even, we know that $f(x)=f(-x)$ Therefore $$\lim_{x\to x_o} x^2=\lim_{x\to -x_o} x^2=L$$
So the limit is the same at the point $x_0$ and $-x_0$. The domain of $f$ is all of $\Bbb{R^1}$. Therefore we need only to look at the interval $I=(0,\infty)$. Now it must be the case that either $x_0 \in I$ or $-x_0 \in I$, assuming $x_0 \neq 0$. $L$ exists if for every $\epsilon >0$ there is a $\delta >0$ : $$\vert x^2 - x_0^2 \vert < \epsilon:0<\vert x - x_0\vert<\delta$$
We can write equivalently that $$-\epsilon < x^2 - x_o^2 < \epsilon$$ $$-\epsilon < (x-x_o)(x+x_o) < \epsilon$$ $$\frac{-\epsilon}{x+x_o} < x - x_o < \frac{\epsilon}{x+x_0}$$
On the interval $I$, we have that $x+x_0>0$. Let $\delta = \frac{\epsilon}{x+x_0}$. As for the case where $x_0=0$, I think we can choose $\delta = \sqrt{\epsilon}$. I don't know how to finish the proof from here.
