If $y=f(x)$ such that $(x+1)^3+(y+1)^3=16$ then evaluate : $\frac{d}{dx}\left(f(x+g(x)).g(x+f(x))\right)$ at $x=-1+\sqrt[3]{15}$

Question

Let $f:\mathbb{R}\to\mathbb{R};y=f(x)$ be an explicit function defined by the implicit equation $$(x+1)^3+(y+1)^3=16$$ Let $g$ be the inverse of $f$.

Evaluate : $$\frac{d}{dx}\left(f(x+g(x)).g(x+f(x))\right)$$ at $x=-1+\sqrt[3]{15}$

My Attempt

If we interchange $x$ and $y$ in the equation of the curve we find that the equation remains unchanged. So,the function f is its own inverse. But beyond this I am not able to do. Should I replace $g(x)$ by $f(x)$

Answer 1

Since $x$ and $y$ are symmetric, so $f(x)=g(x)$. Let $x_0=-1+\sqrt[3]{15}$ and then $y_0=f(x_0)=0$. Thus \begin{eqnarray} &&\frac{d}{dx}\left(f(x+g(x)).g(x+f(x))\right)\bigg|_{x=x_0}\\ &=&2f(x+f(x))f'(x+f(x))(1+f'(x))\bigg|_{x=x_0}\\ &=&2f(x_0+f(x_0))f'(x_0+f(x_0))(1+f'(x_0))=0 \end{eqnarray} since $$ f(x_0+f(x_0))=f(x_0)=0. $$