Answer
$\frac{1}{2}$
Work Step by Step
Using the derivative of inverse function it follows:
$$(f^{-1})'(1)=\frac{1}{f'(a)}$$ where $1=f(a)$
Find $a$.
$$a+a^{2}+e^{a}=1$$ using a graphing calculator it follows that $a\approx -1.515$ and $a=0$
Take the exact value of $a$ it follows:
$$(f^{-1})'(1)=\frac{1}{f'(0)}$$
$$f'(x)=1+2x+e^{x}$$
$$f'(0)=1+2\cdot 0+e^{0}=2$$
Therefore:
$$(f^{-1})'(1)=\frac{1}{2}$$
