Your case is trickier than the standard interchange of partial derivatives. So let us first recall how that works. If you have a function $F: \mathbb{R}^2 \to \mathbb{R}$ which has continuous second partial derivatives then it is true that $$\dfrac{\partial}{\partial x}\dfrac{\partial F (x,y)}{\partial y}=\dfrac{\partial}{\partial y}\dfrac{\partial F(x,y)}{\partial x}$$
However your case is different because you have a 1-variable function $f: \mathbb{R} \to \mathbb{R}$ and another 2-variable function $u:\mathbb{R}^2 \to \mathbb{R}$.
Notice that since $f$ is a one variable function we first need to make sense of the expression $$\dfrac{d}{dx}\dfrac{df}{du}$$ This one is easy to make sense of: By $\dfrac{d}{dx}\dfrac{df}{du}$ we actually mean:
$$\dfrac{d}{dx}\dfrac{df (u(x,t))}{du}$$
So you first take the partial derivative with respect to $u$, then evaluate $u$ at $u(x,t)$ and then take the derivative with respect to $x$.
On the other hand, by the expression $\dfrac{d}{du}\dfrac{df}{dx}$ we actually mean:
$$\dfrac{d}{du}\dfrac{d}{dx}|_{x=x(u,t)}f(u(x,t))$$
i.e., we first plug in $u(x,t)$ in $f$, so now you have a function of $x$ (and $t$), then we differentiate with respect to $x$, then evaluate $x$ at $x=x(u,t)$ (you can do this under some mild assumptions by implicit function theorem) so now we have a function of $u$ and $t$, and then we differentiate with respect to $u$.
As you can see, even though the expressions $\dfrac{d}{dx}\dfrac{df}{du}$ and $\dfrac{d}{du}\dfrac{df}{dx}$ may seem to differ just by the exchange of partial derivatives they are actually quite different expressions since we have a lot of compositions going on in-between. I know of no general conditions that make this equality true and I doubt there is any. This is really quite apparent in your own answer since $f=u^2$ and $u=\sqrt(x)$ are $C^{\infty}$ (where $x>0$, say) and you pretty much can't get regularity better than that, yet it doesn't work on them.
