$\varepsilon-\delta$ proof of this multivariable limit We've to prove that
$$
\lim_{(x,y)\to(0,0)} \frac{x^3+y^4}{x^2+y^2} =0
$$
Kindly check if my proof below is correct.
Proof
We need to show there exists $\delta>0$ for an $\varepsilon>0$ such that
$$
\left| \frac{x^3+y^4}{x^2+y^2} \right| < \varepsilon \implies \sqrt{x^2+y^2}< \delta
$$
Start
$$
\left| \frac{x^3}{x^2+y^2} \right| <\left| \frac{x^3+y^4}{x^2+y^2} \right| < \varepsilon
$$
Note
$$
\left| \frac{x^3}{x^2+y^2} \right|= \frac{x^2|x|}{x^2+y^2}>\frac{(x^2-y^2)|x|}{x^2+y^2}
$$
Therefore
$$
\frac{x^2-y^2}{x^2+y^2}|x|<\varepsilon \tag{1}
$$
Note
$$
\frac{x^2-y^2}{x^2+y^2}|x|<|x|=\sqrt{x^2}<\sqrt{x^2+y^2}<\delta
$$
So
$$
\frac{x^2-y^2}{x^2+y^2}|x|<\delta \tag{2}
$$
From $(1)$ and $(2)$, we can say
$$
\delta=\varepsilon
$$

There is a simple strategy for dealing with such limits, which is just switching to polar coordinates. By setting $x=\rho\cos\theta, y=\rho\sin\theta$ we have $x^2+y^2=\rho^2$ and
$$ \left|\frac{x^3+y^4}{x^2+y^2}\right| = \rho\left|\cos^3\theta+\rho \sin^4\theta\right|. $$
The RHS is trivially bounded by $\rho(1+\rho)$, which is $\leq 2\rho$ if $\rho\leq 1$.
It follows that as long as $\rho\leq\frac{\varepsilon}{2}\leq 1$ we have $|f(x,y)|\leq \varepsilon$.

If you want a sharp bound you may consider that for any sufficiently small $\rho\in\mathbb{R}^+$ the maximum of $|\cos^3\theta+\rho\sin^4\theta|$ is indeed $1$, so $\rho\leq\varepsilon$ is sufficient to ensure $|f(x,y)|\leq\varepsilon$.
