Answer
$f^{58}(0)=0$
Work Step by Step
Maclaurin series of $f(x)=\Sigma_{n=0}^{\infty}\frac{f^{n}(0)x^{n}}{n!}$
Coefficient of $x^{58}=\frac{f^{58}(0)}{n!}$
Note that the powers of $x$ in the expansion of $(1+x^{3})^{30}$ must be multiples of $58$.
But $58$ is not multiple of $3$.
Thus, $0=\frac{f^{58}(0)}{n!}$
Hence, $f^{58}(0)=0$