Proving continuity of the function $ f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto \frac{x^{3}}{1+x^{2}} $ Using only the epsilon delta criterion, show that the function $ f: \mathbb{R} \rightarrow \mathbb{R}, x \mapsto \frac{x^{3}}{1+x^{2}} $ is continuous over its entire domain of definition.
Definition of the epsilon delta criterion:
A function $f$ from $\Bbb R$ to $\Bbb R$ is continuous at a point $x_0 \in \mathbb{R} $ if given $ε > 0$ there exists $δ > 0$ such that if $|x-x_0| < \delta $ then $|f(x)-f(x_0)|< \epsilon$.
I have already started looking for delta, but get stuck here, and don't know how to further transform or estimate to find the appropriate delta:
$ \begin{aligned}\left|f(x)-f\left(x_{0}\right)\right| &=\left|\frac{x^{3}}{1+x^{2}}-\frac{x_{0}^{3}}{1+x_{0}^{2}}\right|=\\ &=\left|\frac{x^{3}\left(1+x_{0}^{2}\right)}{\left(1+x^{2}\right)\left(1+x_{0}^{2}\right)}-\frac{x_{0}^{3}\left(1+x^{2}\right)}{\left(1+x_{0}^{2}\right)\left(1+x^{2}\right)}\right|=\\ &=\left|\frac{x^{3}\left(1+x_{0}^{2}\right)-x_{0}^{3}\left(1+x^{2}\right)}{\left(1+x^{2}\right)\left(1+x_{0}^{2}\right)}\right|=?\end{aligned} $

We have
\begin{align}
|x^3(1+y^2)- y^3(1+x^2)| & = |(x-y)(x^2+x y + y^2 + x^2 y^2)|\\
&\le |x-y|(x^2 + \frac{1}{2}(x^2+y^2) + y^2 + x^2 y^2)\\
&\le \frac{3}{2}|x-y|(x^2 + y^2 + x^2 y^2)\\
&\le \frac{3}{2}|x-y|(1 + x^2)(1+y^2)\\
\end{align}
Hence
\begin{equation}
\frac{|x^3(1+y^2)- y^3(1+x^2)|}{(1 + x^2)(1+y^2)}
\le \frac{\frac{3}{2}|x-y|(1 + x^2)(1+y^2)}{(1 + x^2)(1+y^2)}
\le\frac{3}{2}|x-y|
\end{equation}
Hence
\begin{equation}
|f(x)-f(x_0)|\le \frac{3}{2} |x-x_0|
\end{equation}
