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}$$
Therefore
\begin{align}
\left\lvert f(x,y)-f(x_0,y_0) \right\rvert
& = \left\lvert \frac{y(y-x^2)}{x^4}\right\rvert \\
& = \frac{\left\lvert y \right\rvert \left\lvert y - x^2 \right\rvert}{\left\lvert x \right\rvert^4} \\
& \leq \frac{ \left(\frac{3 y_0 }{2} \right) \left( 2+2 x_0 \right)\varepsilon' }{\frac{ x_0^4 }{16}} \\
& \leq \frac{48(1+x_0)y_0}{x_0^4} \cdot \frac{x_0^4 \cdot \varepsilon}{48(1+x_0)y_0} \\
& = \varepsilon
\end{align}
Therefore $f(x,y)$ is continuous at $(x_0, y_0)$ where $x_0 \gt 0$ and $y_0=x_0^2.$
The case $x_0 \lt 0$ and $y_0=x_0^2$ can be dealt with similarly.
Best Answer
Let's finish it based on your ideas.
As you said, choosing $\delta\leq1$ is a good idea. Notice that it gives us that
$$\lvert x^2\rvert< 1,$$ $$\lvert x-4\rvert\geq\lvert \lvert x\rvert-4\rvert=4-\lvert x\rvert>3.$$
Using this we have that
$$\left\lvert\frac{x^3}{x-4}\right\rvert<\frac{\lvert x\rvert }{3}<\lvert x\rvert.$$
Clearly, as you guessed yourself, $\delta=\varepsilon$ works wonders here. Thus, with
$$\delta=\min\left\{1,\varepsilon\right\}$$
we are done.