derivative of $\frac 2x \sin(x^3)$ by definition The function: $$\frac 2x \sin(x^3)$$ when $x\neq0$, and $0$ while $x=0$.
I need to find if the function is derivation at $x=1$.
First step was to check if the function is continuous. there is just 1 side of limit to check (since its the same function around $x=1$, so I compared the limit of the function with $f(1)$:
$$\lim_{x\to 1}\space \frac{2}{x}\sin(x^3) = 2\sin(1) = f(1)$$
first question: was this step necessary?
next:
$$f'(1) =\lim_{h\to 0}\space \frac{ \frac2{h+1}\sin((h+1)^3)-2\sin(1)}h$$
$$=\lim_{h\to 0}\space \frac{ \frac{2\sin((h+1)^3)}{h+1}-2\sin(1)}h$$
and I know about $\lim_{x\to 0}\frac {\sin(x)}x = 1$, so:
$$=\lim_{h\to 0}\space \frac{ \frac{2\sin((h+1)^3) \cdot (h+1)^2}{(h+1)^3}-2\sin(1)}h$$
and then:
$$=\lim_{h\to 0}\space \frac{ 2(h+1)^2-2\sin(1)}h$$
and now I don't know how to get rid of the $h$ denominator

I did not answer your exact question, I looked at the case of $x=0$ rather than $x=1$, but the argument is analogous. 
Evaluate the derivative of $f : \mathbb{R} \to \mathbb{R}$, where $$f(x) : = \begin{cases}
\frac{2}{x} \sin(x^3), & x \in \mathbb{R} \backslash \{ 0 \}, \\
0, & \text{otherwise}.
\end{cases}$$ 
Let us first verify that $f$ is at least continuous at $x=0$, let alone differentiable. To this end, write $\sin(x)$ as its Taylor expansion $$\sin(x) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} x^{2k+1}.$$ Then $$\frac{2}{x} \sin(x^3) = 2 \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} x^{3(2k+1) - 1} = 2\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} x^{6k+2}. $$ We therefore see that $$\lim_{x \to 0} \frac{2}{x} \sin(x^3) =0,$$ verifying that $f$ is continuous at $0$. Now let us check whether $f$ is differentiable at $x=0$. Indeed, we observe that \begin{eqnarray*}
\lim_{\Delta x \to 0} \frac{f(0+\Delta x) - f(0)}{\Delta x} &=& \lim_{\Delta x \to 0} \frac{f(\Delta x) - f(0)}{\Delta x} \\
&=& \lim_{\Delta x \to 0} \frac{f(\Delta x)}{\Delta x} \\
&=& \lim_{\Delta x \to 0} \frac{2}{\Delta x \cdot \Delta x} \sin ( (\Delta x)^3 ). 
\end{eqnarray*} For simplicity, let $z = \Delta x$ and note that we just need to determine whether the limit $$\lim_{z \to 0} \frac{2 \sin(z^3)}{z^2}$$ exists. Analogously to the above continuity argument, write $\sin(z^3)$ as its Taylor expansion, $$\sin(z^3) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} z^{3(2k+1)}.$$ It then follows that \begin{eqnarray*}
\lim_{z \to 0} \frac{2}{z^2} \sin(z^3) &=& \lim_{z \to 0} 2 \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} z^{6k+2} \\
&=& 0,
\end{eqnarray*} so $f$ is differentiable with $f'(0)=0$. 
