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

The function is obviously continuous at $1$, because it is a composition of continuous functions.
Your computation is flawed: it is true that $\lim_{x\to0}\frac{\sin x}{x}=1$, but you have $\lim_{h\to0}\frac{\sin((1+h)^3)}{(1+h)^3}$ and the preceding limit doesn't apply.
For the derivative at $1$:
$$
f'(1)=\lim_{h\to0}\frac{\dfrac{2\sin((1+h)^3)}{1+h}-2\sin1}{h}
$$
Now write $(1+h)^3=1+hg(h)$, where $g(h)=3+3h+h^2$, so you have to compute, disregarding the factor $2$,
$$
\frac{\sin(1+hg(h))-(1+h)\sin 1}{h(1+h)}=
\frac{\sin1\cos(hg(h))+\cos1\sin(hg(h))-\sin 1-h\sin 1}{h(1+h)}
$$
The right hand side can be rewritten as
$$
\frac{\cos(hg(h))-1}{hg(h)}\frac{g(h)\sin 1}{1+h}
+\frac{g(h)\cos 1}{1+h}\frac{\sin(hg(h))}{hg(h)}
-\frac{\sin 1}{1+h}
$$
Now the limit is basically done; the first fraction has limit $0$, the second fraction has limit $3\sin 1$; the third fraction has limit $3\cos 1$, the fourth fraction has limit $1$; the fifth fraction has limit $\sin 1$. Thus, reinserting the factor $2$:
$$
f'(1)=2(3\cos1-\sin1)
$$
Just to check with the standard procedure: for $x\ne0$,
$$
f'(x)=2\frac{3x^3\cos(x^3)-\sin(x^3)}{x^2}
$$
and $f'(1)=2(3\cos 1-\sin 1)$.
Are you sure you don't want to compute the derivative at $0$? In order to do it, you need to first check the function is continous at $0$:
$$
\lim_{x\to0}\frac{2\sin(x^3)}{x}=
\lim_{x\to0}2x^2\frac{\sin(x^3)}{x^3}=0
$$
Then
$$
f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}
=\lim_{h\to0}\frac{2\sin(h^3)}{h^2}
=\lim_{h\to0}2h\frac{\sin(h^3)}{h^3}
$$
