You have gotten to \(\displaystyle \displaystyle \begin{align*} \left| x - x_0 \right| \left| x^2 + x\, x_0 + x_0 ^2 \right| < \epsilon \end{align*}\). Great job. From this we have \(\displaystyle \displaystyle \begin{align*} \left| x - x_0 \right| < \frac{\epsilon}{\left| x^2 + x\, x_0 + x_0^2 \right|} \end{align*}\). We need this denominator to be at its maximum so that the fraction on the right will be at its minimum, to give us a value we can set for \(\displaystyle \displaystyle \begin{align*} \delta \end{align*}\). To maximise the denominator, we need an upper bound, so we can use the triangle inequality...

\(\displaystyle \displaystyle \begin{align*} \left| x^2 + x\,x_0 + x_0^2 \right| &\leq \left|x \right|^2 + \left| x \right| \left|x_0 \right| + \left| x_0 \right|^2 \textrm{ by the Triangle Inequality } \end{align*}\)

Unfortunately, this denominator is unconstrained, but no matter, we only care about what's happening NEAR \(\displaystyle \displaystyle \begin{align*} x = x_0 \end{align*}\), so we restrict the distance between \(\displaystyle \displaystyle \begin{align*} x \end{align*}\) and \(\displaystyle \displaystyle \begin{align*} x_0 \end{align*}\) to within some small value, say 1. Then \(\displaystyle \displaystyle \begin{align*} \left| x - x_0 \right| < 1 \end{align*}\).

The first thing we can determine from \(\displaystyle \displaystyle \begin{align*} \left| x - x_0 \right| < 1 \end{align*}\) is that

\(\displaystyle \displaystyle \begin{align*} -1 < x - x_0 &< 1 \\ -1 + x_0 < x &< 1 + x_0 \\ |x| &< 1 + x_0 \end{align*}\)

So that means \(\displaystyle \displaystyle \begin{align*} \left| x \right|^2 + \left| x \right| \left|x_0 \right| + \left| x_0 \right|^2 &< \left( 1 + x_0 \right) ^2 + \left( 1 + x_0 \right) \left| x_0 \right| + \left| x_0 \right| ^2 \end{align*}\)

Therefore we can say \(\displaystyle \displaystyle \begin{align*} \frac{\epsilon}{\left| x^2 + x\,x_0 + x_0^2 \right|} \end{align*}\) is minimised at \(\displaystyle \displaystyle \begin{align*} \frac{\epsilon}{ \left( 1 + x_0 \right)^2 + \left(1 + x_0 \right)\left| x_0 \right| + \left|x _0 \right|^2} \end{align*}\).

So finally we can let \(\displaystyle \displaystyle \begin{align*} \delta = \min \left\{ 1, \frac{\epsilon}{ \left( 1 + x_0 \right)^2 + \left( 1 + x_0 \right) \left| x_0 \right| + \left| x_0 \right|^2 } \right\} \end{align*}\) and reverse each step to complete your proof.