Answer
$$y''=-\frac{5 x^{4}}{y^{11}}$$
Work Step by Step
Given
$$x^6+y^6=1$$
Differentiate with respect to $x$
$$6x^5+6y^5y'=0\ \ \Rightarrow \ \ y'=\frac{-x^5}{y^5}$$
and
\begin{align*}
y^{\prime \prime}&=-\frac{y^{5}\left(5 x^{4}\right)-x^{5}\left(5 y^{4} y^{\prime}\right)}{\left(y^{5}\right)^{2}}\\
&=-\frac{5 x^{4} y^{4}\left[y-x\left(-x^{5} / y^{5}\right)\right]}{y^{10}}\\
&=-\frac{5 x^{4}\left[\left(y^{6}+x^{6}\right) / y^{5}\right]}{y^{6}}\\
&=-\frac{5 x^{4}}{y^{11}}
\end{align*}
