I could simplify the problem quite a bit but left a simpler problem for you to solve. We have
$$f(x) = \left(e^{\frac{1}{x}}-1\right) x^2-\frac{x^2}{x+1}$$
using some basic algebra you can factor its derivate to:
$$f'(x) = \frac{2 e^{\frac{1}{x}} x^3-2 x^3+3 e^{\frac{1}{x}} x^2-5 x^2-4 x-e^{\frac{1}{x}}}{(x+1)^2}$$
So what is left to show is that
$$h(x)= e^{\frac{1}{x}} x^3-2 x^3+3 e^{\frac{1}{x}} x^2-5 x^2-4 x-e^{\frac{1}{x}}$$
has exactly one real root in $(0, \infty)$, but this is not hard because if we look at the graph of h we see that

And therefore assume that $$h'(x)=(x+1) \left(e^{\frac{1}{x}} \left(\frac{1}{x^2}+6 x-\frac{1}{x}-2\right)-6 x-4\right)>0$$
Where you can also use the notation
$$h'(t)=h'(1/x)=\frac{(t+1) \left(-4 t+e^t ((t-2) t (t+1)+6)-6\right)}{t^2}$$
So it remains again to show that
$$ (t+1) \left(t^3-t^2-2 t+6\right) e^t\geq 2 \left(2 t^2+5 t+3\right) $$
Then we are done if we apply the mean value theorem. It should be doable to show this.