Find $y$-Lipschitz constant $$f(x,y)=x^3e^{-xy^2}, 0\leq x\leq a, y\in \mathbb R, a>0$$
I need to find $K>0$ such that $$|f(x,y_1)-f(x, y_2)|\leq K|y_1-y_2|$$ for all $0\leq x\leq a$ and $y_1,y_2\in \mathbb R$
I did this $$|x^3e^{-xy_1^2}-x^3e^{-xy_2^2}|=x^3|e^{-xy_1^2}-e^{-xy_2^2}|\leq a^3|e^{-xy_1^2}-e^{-xy_2^2}|$$
now I don't know what to do.
 A: Better to find $\frac{df}{dy}$, then you will get either $2a^3$  or $2a^4$.
A: Lipschitz constant Lipschitz constant Lipschitz constant Lipschitz constant

A: Try to find a Lipschitz constant of $x \mapsto \mathrm{e}^{-x}$, $x \ge 0$. Hint: Fundamental Theorem of Calculus.
Then, you can use this in your Lipschitz estimate.
A: I'll go on from where you stopped at $ a^3|e^{-xy_1^2}-e^{-xy_2^2}|$. Let's note that $ 0 \leq e^{-xy_1^2}\leq 1$ because this is a monotone decreasing function of $x$, and therefore, it receives a maximum when $x=0$, such that:
$ a^3|e^{-xy_1^2}-e^{-xy_2^2}|\leq a^3|1-e^{-xy_2^2}|.$
Let's use the fact that $ e^{-xy_2^2}\geq 0$, to conclude that:
$ a^3|e^{-xy_1^2}-e^{-xy_2^2}|\leq a^3|1-e^{-xy_2^2}|\leq a^3\cdot 1=a^3$
A: The Mean Value Theorem says that the $y$-Lipschitz constant would be
$$
\sup_{\substack{x\in[0,a]\\y\in\mathbb{R}}}\left|\frac{\partial}{\partial y}f(x,y)\right|
$$
So we compute
$$
\begin{align}
\frac{\partial}{\partial y}f(x,y)
&=-2x^4ye^{-xy^2}\\
&=-2x^{7/2}x^{1/2}ye^{-xy^2}\\
&=-2x^{7/2}ue^{-u^2}\\
\left|\frac{\partial}{\partial y}f(x,y)\right|&\le2a^{7/2}\frac1{\sqrt{2e}}
\end{align}
$$
where we find the maximum of $ue^{-u^2}$ by taking its derivative and solving
$$
(1-2u^2)e^{-u^2}=0
$$
to get $u=1/\sqrt2$. Thus, the maximum of $ue^{-u^2}$ is $1/\sqrt{2e}$.
Therefore,
$$
|f(x,y_1)-f(x, y_2)|\le a^{7/2}\sqrt{2/e}\,|y_1-y_2|
$$
