No, it's not! Consider for example
$f(x,y) := g(y) \cdot \left( \frac{1}{\sqrt{2\pi}} \cdot \exp \left(- \frac{x^2}{2} \right) \right)^{\frac{1}{2}}$
where $g \in L^1$ is an arbritary function. Then $\int_{-\infty}^\infty |f(x,y)|^2 \, dx = |g(y)|^2 \cdot \int_{\mathbb{R}}\frac{1}{\sqrt{2\pi}} \cdot \exp \left(- \frac{x^2}{2} \right) \, dx = |g(y)|^2$ since the (remaining) integrand is the density of the normal distribution $N(0,1)$.
You could choose for example $g(y) := e^{-{y^2}}$, then
$\frac{d}{dy} \int_{-\infty}^\infty |f(x,y)|^2 \, dx = \frac{d}{dy} e^{-2y^2} = -4y \cdot e^{-2y^2} \not= 0$
The equality $\frac{d}{dy} \int_{-\infty}^\infty |f(x,y)|^2 \, dx=0$ holds iff $\int_{-\infty}^\infty |f(x,y)|^2 \, dx =$ constant.