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Why can't I interchange integration and differentiation here?

I think my questions relates to this other: "counterexamples to differentiation under integral sign"

In fact, it provides a counterexample

Consider $f(x,y)=y^3e^{-y^2x}$ and define $F(y) =\int_0^{\infty}f(x,y)dx$

We have that $F'(0)\not = \int_0^{\infty} \frac{\partial f}{\partial y}(x,0)dx$

(We calculate $F'(0)$ essentially using Monotone convergence theorem we can show that, for $y\in \mathbb{R}\setminus\{0\} $, $F(y)=y$ moreover $F(0)=0$ so $F'(0)=1$)

Now, I want to understand which hypothesis of Theorem 2 at this page does not hold. Obviously the third hypothesis does not hold, but I want to consider the case in which we replace it by a weacker condition:

"For each $b \in \mathbb{R}$, there exists an open interval $b\in J$ and an integrable function $g(x)$ over $(0, \infty)$ ,such that $| \frac{\partial f}{\partial y}(x,y)| \leq g(x)$ for every $y\in J$ and $\forall x$"

Now, the first hypothesis certainly holds as $\forall y, \ x\rightarrow f(x,y)$ is integrable $(0,\infty)$ by comparison with $e^{-kx}$ for appropriate positive value of $k$

Moreover $ \frac{\partial f}{\partial y}(x,y)$ exists everywhere...

So is the last hypothesis to be problematic but I can't see how as I can bound $y$ in $J$ and then just use some linear combination of $e^{-kx}$ and $ xe^{-lx}$ for suitable $k,l$ as they are both integrable over $(0,\infty)$

Thank you very much!