That looks pretty straight forward to me. Obviously when n= 0 or 1, $\displaystyle f(0)= 0< 1= 5^0$ and $\displaystyle f(1)= 1< 5= 5^1$.
Now, suppose that $\displaystyle f(k)< 5^k$ for all k< n. Then $\displaystyle f(n+1)= 3f(n)+ 10f(n-1)< 3(5^n)+ 10(5^{n-1})$.
Can you simplify that?