There is no "sure fire" way of proving continuity of a function. However, the steps are usually a bit backward to what the actual definition is. That is, the definition says that $f$ is continuous at $a$ if for each $\epsilon>0$, there exists $\delta >0$ such that if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$.
We start the proof by taking an arbitrary $\epsilon > 0$. However, we then usually do not magically think of a $\delta$ that would fit. What we typically try to do is simplify the expression $|f(x)-f(a)|$ and prove that it is "small", keeping in the back of our mind that we can always make $|x-a|$ "small".
In your case, for example, we can prove first that $|f(x)-f(a)|=|\frac{x-a}{\sqrt x + \sqrt a}|$. Now, we want to ask ourselves: If $|x-a|$ is small, is this expression also small?
This is done by trying to find a small upper bound for the expression that will hold whenever $|x-a|<\delta$. In your particular case, this is fairly simple, since we know that the expression is smaller than $\frac{|x-a|}{\sqrt a}$, and this is smaller than $\epsilon$ if $\delta$ is set small enough.
Now, once we did these steps, we take a step back, and think about what we just did.
Looks like we found our $\delta$, and we have to set our $\delta=\epsilon\sqrt a$.
Hmm, is this OK? Well, as long as $a>0$, we know that $\delta>0$, and we have proven that if $|x-a|<\delta$, then $|f(x)-f(a)|<\frac{\delta}{\sqrt a}$, ,which is equal to $\epsilon$, so it looks like we are almost done.
Almost, because we can see that if $\sqrt a=0$, then our proof does not work! We need to do that part separately. If $a=0$, then $|f(x)-f(a)|=|\sqrt x|$. We still need to prove that if $|x-0|$ is small, then $|\sqrt x|$ is also small.
Here is a hint which you can play around with to aid your understanding:
Let $\delta > 0$. Notice that if you choose $x$ such that $0 < |x| < \min\left(\delta, \frac{1}{|c|+1}\right)$ then
$$
|f(x) - c| = \left|\frac{1}{x} - c \right| = \left|\frac{1-cx}{x} \right| \geq \frac{1 - \frac{|c|}{|c|+1}}{\frac{1}{|c|+1}} = 1
$$
Best Answer
Case $(1)$
Since $f(x_0,0)=0$ for $x \neq 0$, to prove continuity of $f(x,y)$ at $(x_0,0)$, we need to show that
For any $\varepsilon \gt 0$, there exists $\delta \gt 0$ s.t. $$\sqrt {(x-x_0)^2+y^2} \lt \delta \implies \left\lvert \frac{y(y-x^2)}{x^4}\right\rvert \lt \varepsilon.$$
Notice that if $\color{red}{\delta =\min \left( \frac{\left\lvert x_0 \right\rvert}{2}, \frac{\left\lvert x_0 \right\rvert^2}{4}, \frac{\left\lvert x_0 \right\rvert^2\varepsilon}{40} \right)}$, then
$$\sqrt {(x-x_0)^2+y^2} \lt \delta $$ $$\implies \sqrt {(x-x_0)^2+y^2} \lt \frac{\left\lvert x_0 \right\rvert}{2} $$ $$\implies \left\lvert x-x_0 \right\rvert \lt \frac{\left\lvert x_0 \right\rvert}{2} $$
$$\implies \frac{\left\lvert x_0 \right\rvert}{2} \lt x \lt \frac{3\left\lvert x_0 \right\rvert}{2}$$
We also have $$\sqrt {(x-x_0)^2+y^2} \lt \delta $$ $$\implies \sqrt {(x-x_0)^2+y^2} \lt \frac{\left\lvert x_0 \right\rvert^2}{4}$$ $$ \implies \lvert y\rvert \lt \frac{\left\lvert x_0 \right\rvert^2}{4}$$
Also $$\sqrt {(x-x_0)^2+y^2} \lt \delta $$ $$\implies \sqrt {(x-x_0)^2+y^2} \lt \frac{\left\lvert x_0 \right\rvert^2\varepsilon}{40}$$ $$ \implies \lvert y\rvert \lt \frac{\left\lvert x_0 \right\rvert^2 \varepsilon}{40}$$
Therefore \begin{align} \left\lvert f(x,y)-f(x_0,0) \right\rvert & = \left\lvert \frac{y(y-x^2)}{x^4}\right\rvert \\ & \leq \frac{\left\lvert y \right\rvert (\left\lvert y \right\rvert + \left\lvert x \right\rvert^2)}{\left\lvert x \right\rvert^4} \\ & \leq \frac{ \left(\frac{\left\lvert x_0 \right\rvert^2 \varepsilon}{40} \right) ( \frac{\left\lvert x_0 \right\rvert^2}{4} + \frac{\left\lvert 9x_0 \right\rvert^2}{4})}{\frac{\left\lvert x_0 \right\rvert^4}{16}} \\ & = \varepsilon \end{align}
$\therefore f(x,y)$ is continuous at $(x_0, 0)$ where $x_0 \neq 0.$
$$$$ Case $(2)$ For any point $(x_0, y_0)$ where $y_0=x_0^2$ and $x_0 \neq 0$
Case $(i)$ $x_0 \gt 0$
For any $\varepsilon \gt 0$, take $$\color{red}{\delta = \min \left(\frac{x_0}{2}, \frac{y_0}{2}, \varepsilon' \right)}$$ where $$\color{red}{\varepsilon' = \min \left(1, \frac{x_0}{2}, \frac{y_0}{2}, \frac{x_0^4 \varepsilon}{48(1+x_0)y_0} \right)}$$
Then similar to case $(1)$, when $\sqrt {(x-x_0)^2+y^2} \lt \delta \leq \varepsilon'$, we have
$$x_0-\varepsilon' \lt x \lt x_0+\varepsilon'$$ $$x_0^2-2x_0\varepsilon'+\varepsilon'^2 \lt x^2 \lt x_0^2+2x_0\varepsilon'+\varepsilon'^2 \tag{1}$$
Similarly we have $$y_0-\varepsilon' \lt y \lt y_0+\varepsilon' \tag{2}$$
From $(1), (2)$, we have $$y_0-x_0^2-\varepsilon'\left(1+2x_0+ \varepsilon'\right) \lt y-x^2 \lt y_0-x_0^2+\varepsilon'\left(1+2x_0- \varepsilon'\right)$$
$$-\varepsilon'\left(1+2x_0+ \varepsilon'\right) \lt y-x^2 \lt \varepsilon'\left(1+2x_0- \varepsilon'\right)$$ Since $\varepsilon' \leq 1,$ $$\lvert y-x^2\rvert \lt \varepsilon'\left(2+2x_0 \right) $$
Also note that $$\sqrt {(x-x_0)^2+y^2} \lt \delta \leq \frac{x_0}{2} \implies \frac{x_0}{2} \lt x \lt \frac{3x_0}{2}$$
and $$\sqrt {(x-x_0)^2+y^2} \lt \delta \leq \frac{y_0}{2} \implies \frac{y_0}{2} \lt y \lt \frac{3y_0}{2}$$
Therefore
\begin{align} \left\lvert f(x,y)-f(x_0,y_0) \right\rvert & = \left\lvert \frac{y(y-x^2)}{x^4}\right\rvert \\ & = \frac{\left\lvert y \right\rvert \left\lvert y - x^2 \right\rvert}{\left\lvert x \right\rvert^4} \\ & \leq \frac{ \left(\frac{3 y_0 }{2} \right) \left( 2+2 x_0 \right)\varepsilon' }{\frac{ x_0^4 }{16}} \\ & \leq \frac{48(1+x_0)y_0}{x_0^4} \cdot \frac{x_0^4 \cdot \varepsilon}{48(1+x_0)y_0} \\ & = \varepsilon \end{align}
Therefore $f(x,y)$ is continuous at $(x_0, y_0)$ where $x_0 \gt 0$ and $y_0=x_0^2.$
The case $x_0 \lt 0$ and $y_0=x_0^2$ can be dealt with similarly.