What is wrong with this proof of $0=1$? I am trying to understand what is wrong with the proof posted here that $0=1$ (source):  

Given any $x$, we have (by using the substitution $u=x^2/y$) $$\large\int_0^1\frac{x^3}{y^2}e^{-x^2/y}\,dy=\Biggl[xe^{-x^2/y}\Biggr]_0^1=xe^{-x^2}.$$
  Therefore, for all $x$,
  \begin{align}
e^{-x^2}(1-2x^2)&=\frac{d}{dx}(xe^{-x^2})\\[0.5em]
&= \frac{d}{dx}\int_0^1\frac{x^3}{y^2}e^{-x^2/y}\,dy\\[0.5em]
&= \int_0^1\frac{\partial}{\partial x}\Biggl(\frac{x^3}{y^2}e^{-x^2/y}\Biggr)\,dy\\[0.5em]
&= \int_0^1 e^{-x^2/y}\Biggl(\frac{3x^2}{y^2}-\frac{2x^4}{y^3}\Biggr)\,dy.
\end{align}
  Now set $x=0$; the left-hand side is $e^0(1-0)=1$, but the right-hand side is $\int_0^1 0\,dy=0$. 

Surely it must be incorrect somewhere however I am not sure at which step it fails.
 A: Too long for a comment, but not an answer.
The fractions $1/y$ are a red herring, at least.
Let $u=\frac{1}{y}$ then you have:
$$f(x)=\int_{0}^1\frac{x^3}{y^2}e^{-x^2/y}\,dy = \int_{1}^{\infty} x^3e^{-x^2u}\,du=xe^{-x^2}$$
Then the argument would have:
$$\begin{align}e^{-x^2}(1-2x^2)&=f'(x)\\
&=\frac{\partial}{\partial x} \int_{1}^{\infty} x^3e^{-x^2u}\,du
\\&=\int_{1}^{\infty}\frac{\partial}{\partial x}(x^3e^{-x^2u})\,du
\\&=\int_{1}^{\infty}(3x^2-2x^4u)e^{-x^2u}\,du
\end{align}$$
So the same argument works without the fractions.
A: Continuing @Thomas Andrews nice calculations that simplify understanding, one can notice that there is no problem here with derivation under the integral sign. It is actually true that 
$$
e^{-x^2}(1-2x^2)=\int_1^\infty (3x^2-2x^4u)e^{-x^2u}\,du.
$$
The problem is that it is not possible to set $x=0$. Looking, for example, at the first part of the last integral 
$$
\int_1^\infty x^2e^{-x^2u}\,du
$$
one could believe that setting $x=0$ would give the zero integrand and, hence, the zero integral. Let us see what happens when $x\to 0$. The integrand, indeed, goes to zero uniformly (that would be enough for a finite measure set), however, for small $x$ the tails of the integrand (for large $u$) become heavier and heavier. If we maximize the integrand for a given $u$ we get $e^{-1}/u$ at $x^2=1/u$. It is not an integrable function, so Lebesgue Dominated Convergence theorem cannot be applied. Neither monotone convergence is true here. In fact, the integral goes to one
$$
\int_1^\infty x^2e^{-x^2u}\,du=\int_1^\infty e^{-x^2u}\,d(x^2u)=
-e^{-x^2u}\bigg|_1^\infty=e^{-x^2}\to 1, \quad x\to 0.
$$
A: Let $u=x^2/y$, then $y=x^2/u$:
$$
\int_0^1 e^{-x^2/y}\left(\frac{3x^2}{y^2}-\frac{2x^4}{y^3}\right)\,\mathrm{d}y
=\int_{x^2}^\infty e^{-u}(3-2u)\,\mathrm{d}u
$$
Therefore,
$$
\begin{align}
\lim_{x\to0}\int_0^1 e^{-x^2/y}\left(\frac{3x^2}{y^2}-\frac{2x^4}{y^3}\right)\,\mathrm{d}y
&=\int_0^\infty e^{-u}(3-2u)\,\mathrm{d}u\\[6pt]
&=1
\end{align}
$$
which is not $0$ as we would seem to get by plugging $x=0$ into the integral on the left. That is, the integral on the left is not continuous at $x=0$.
