Showing $\lim_{x\to 0}\frac{\sin(x^2\sin\frac{1}{x})}{x}=0$

I assume you're using that $\sin (r)\approx r$ in your first step. While this is a good heuristic, it does not constitute a rigorous argument.

Perhaps try something like

$$ \left|\frac{\sin(x^2\sin\frac{1}{x})}{x}\right|\le \frac{x^2\left|\sin\frac{1}{x}\right|}{|x|}\le |x| $$ and the RHS tends to $0$ as $x\to 0$.