Let $f(x)$ be a polynomial such that $f(f(x))=f(x-1)f'(x+1)$ and $f'(0)=2$. Find $f(x)$. 
Let $f(x)$ be a polynomial such that $f(f(x))=f(x-1)f'(x+1)$ and $f'(0)=2$. Find $f(x)$.

Putting $x=-1$, I get $f(f(-1))=f(-2)f'(0)=2f(-2)$
Putting $x=0$, I get $f(f(0))=f(-1)f'(1)$
Also, replacing $x$ by $x+1$, I get $f(f(x+1))=f(x)f'(x+2)$
Not sure if I am getting anywhere.

From the hint given by @NinadMunshi and @VIVID, I am able to solve the question. Here is the solution:
Degree of $f(f(x))$= degree of $f(x-1)f'(x+1)\implies n^2=n+n-1$, $n$ being the degree of $f(x)$.
So, $n=1\implies  f(x)=ax+b\implies f'(x)=a\implies a=2\implies f(x)=2x+b$
Putting this in $f(f(x))=f(x-1)f'(x+1)$, I get $2(2x+b)+b=(2(x-1)+b)\cdot2\implies b=-4$
So, $f(x)=2x-4$
