Well, I am a freshman, This was my homework. With a bit of googling, I found that: $$f^{-1}(x) = \pm \frac{\sqrt{W(2x^2)}}{\sqrt{2}}$$ What our professor did was: $$ f(x)=xe^{x^{2}}=x+x^3+\dfrac{x^5}{2} +O(x^6) $$ Then he proved that $f(x)$ is a bijection then he supposed that: $$f^{-1}(x) = a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+O(x^6)$$ then $$ f^{-1}(f(x))=x $$ $$ f^{-1}(f(x))=a_0+a_1{(x+x^3+\dfrac{x^5}{2})}+a_2{(x+x^3+\dfrac{x^5}{2})}^2+a_3{(x+x^3+\dfrac{x^5}{2})}^3+a_4{(x+x^3+\dfrac{x^5}{2})}^4+a_5{(x+x^3+\dfrac{x^5}{2})}^5 $$

then solved the following for the coefficient: $$ a_0+a_1{(x+x^3+\dfrac{x^5}{2})}+a_2{(x+x^3+\dfrac{x^5}{2})}^2+a_3{(x+x^3+\dfrac{x^5}{2})}^3+a_4{(x+x^3+\dfrac{x^5}{2})}^4+a_5{(x+x^3+\dfrac{x^5}{2})}^5=x $$ and found that: $$ a_0=0,a_1=1,a_2=0,a_3=-1,a_5=\dfrac{5}{2} $$ and found that: $$f^{-1}(x)=x-x^3+\dfrac{5x^5}{2} +O(x^6)$$ I Think it is wrong because when you plot both of them, they are not symmetric to $y = x$


$$\text{I don't agree with} \quad f^{-1}(x)=x-x^3-\dfrac{1}{2}x^5 +O(x^6)$$ I found : $$f^{-1}(x)=x-x^3+\dfrac{5}{2}x^5 +O(x^6)$$

But my answer is mainly about the question of symmetry to $y=x$.

Sure, you are right. The curves $y(x)$ and $f^{-1}(x)$ must be symmetrical.

The series $y=x-x^3+\dfrac{5}{2}x^5$ is an approximate of $f^{-1}(x)$ only accurate in a range close to the origin. So, don't expect to observe symmetrical curves if you consider a too large range from the origin.

This is illustrated on the figure below.

$y=x$ is drawn in black.

$x\,e^{x^2}$ is drawn in red (solid curve).

$f^{-1}(x) $ is drawn in red (dashed curve), that is the symmetric curve of $x\,e^{x^2}$.

$x-x^3+\dfrac{5}{2}x^5$ is the first blue curve. One can see that the approximation is correct only on a small range, approximately $x<0.3$

More accuracy requires more terms for the series. For example : $x-x^3+\dfrac{5}{2}x^5-\dfrac{49}{6}x^7+\dfrac{243}{8}x^9-\dfrac{14641}{120}x^{11}$ is the second blue curve.

One can see that the range is not much enlarged, approximately $x<0.4$ . This is only a small improvement. Definitively one cannot expect a large range of validity for the inverse function with this method of limited series expansion, even with a large number of terms.

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