Proving $\frac{d}{dx}x^2=2x$ by definition
Yes it is correct. A tip is to immediately write: $$\lim_{h \rightarrow 0} \frac{h^2+2xh}{h}=\lim_{h \rightarrow 0} 2x+h.$$ You can now see directly, to get $2x+h$ within $\epsilon>0$ of $2x$, we must choose $|h|<\epsilon$.
Note that if $f$ is a continuous function, and $f(0)$ exists, then $\lim_{h\to 0} f(h) = f(0)$ (this is just the definition of continuity).
This lets you immediately plug in $h=0$ in after dividing by $h$, avoiding the epsilon-delta portion of your argument.