Show that $f(x) = xe^{x^2}$ is invertible and determine the 6th degree maclaurin polynomial of $f^{-1}(x)$ Show that $f(x) = xe^{x^2}$ is invertible and determine the 6th degree maclaurin polynomial of $f^{-1}(x)$.
I can see that $f$ is invertible since $x$ is strictly increasing and when $x < 0$ $e^{x^2}$ is strictly decreasing and when $x > 0$ $e^{x^2}$ is strictly increasing.  However, I don't know how to find $f^{-1}$? I've tried by the usual approach of setting $xe^{x^2} = y$ and solve for $x$ but I don't get anywhere. Once I've found the inverse the maclaurin polynomial should be straight forward I think.

It is not necessary calculate $f^{-1}$ analytically because the derivative of $f^{-1}$ can be written in function of the derivative of $f$, infact you have that 
$(f^{-1})’(y)=\frac{1}{f’(f^{-1}(y))}$
You can calculate the other derivatives of $f^{-1}$ starting from this relation. 
In your case you have that 
$f(0)=0$ 
So $f^{-1}(0)=0$ that means that 
$(f^{-1})’(0)=\frac{1}{f’(0)}$
For the second derivative, you have that 
$(f^{-1})’’(y)=(\frac{1}{f’(f^{-1}(y))})’=-\frac{1}{(f’(f^{-1}(y))^2} (f’(f^{-1}(y))’ = $
$=-\frac{f’’(f^{-1}(y))}{(f’(f^{-1}(y))^3} $
So
$(f^{-1})’’(0)= -\frac{f’’(0)}{(f’(0))^3}$
You can understand that this is not a good computable method to calculate the Maclaurin expansion because you must do a lot of calculus. In any case you can expand the function $f$ in Maclaurin polynomial of order $6$ and you can calculate the coefficients of $f^{-1}$ if you impose the following condition: 
$f^{-1}(f(x))=x$
