$f_n(x)$ = $nx(1-x^4)^n$ for $x\in [0,1]$
$f(x)$ = $0$, clearly.
For uniform convergence: After calculating the derivative, I found out the the maximum of $|f_n(x)|$ is at $x= (\frac{1}{1+4n})^{1/4}$
${\rm max }\ |f_n(x)|\rightarrow 0$.
Is that right? Im not sure about the limit of the maximum.
Taking $x_n = n^{-1/4} \in [0,1]$, we have
$$f_n(x_n) = n n^{-1/4}(1-(n^{-1/4})^4)^n = n^{3/4}(1 - 1/n)^n,$$
and
$$ \lim_{n \to \infty} \sup_{x \in [0,1]}|f_n(x)| \geqslant\lim_{n \to \infty}f_n(x_n) = \infty \cdot e^{-1} = \infty$$
Thus, convergence is non-uniform.