Given any $x$, we have (by using the substitution $u=x^2/y$)
$$ \int_0^1 {x^3\over y^2} e^{-x^2/y}\,dy
  = \biggl[x e^{-x^2/y}\biggr]_0^1 = x e^{-x^2}.$$
Therefore, for all $x$,
$$\eqalign{e^{-x^2}(1-2x^2) &= {d\over dx}(xe^{-x^2})\cr
 &= {d\over dx} \int_0^1 {x^3\over y^2} e^{-x^2/y}\,dy\cr
 &= \int_0^1 {\partial \over \partial x} \biggl({x^3\over y^2}
       e^{-x^2/y}\biggr)\,dy\cr
 &= \int_0^1 e^{-x^2/y} \biggl({3x^2\over y^2} -
       {2x^4\over y^3}\biggr)\,dy.\cr}
$$
Now set $x=0$; the left-hand side is $e^0(1-0) = 1$,
but the right-hand side is $\int_0^1 0\,dy = 0$.