I know FT{$\frac{\partial u}{\partial x}$} = (ik)FT{u}. Give a function $U(x,y)$. Is the following true?
FT{ $\frac{\partial^2 U}{\partial y \partial x}$} = FT{$\frac{\partial U}{\partial y}$} FT{$\frac{\partial U}{\partial x}$} = (ik_y)(ik_x FS{U}) = (ik_y k_x FS{U}? If not, what does it equal to?
FT = Fourier Transform
On
If you start with variables $x,y,z$, and your F.T. variables are $\xi_1,\xi_2,\xi_3$, then
$$\widehat{\frac{\partial^2 U}{\partial x \partial x}} = \widehat{\frac{\partial}{\partial x}\frac{\partial U}{\partial x}} = -i\xi_1 \widehat{\frac{\partial U}{\partial x}} = i^2 \xi_1^2 \widehat U = -\xi_1^2 \widehat U$$
(up to some factors of $2 \pi$, depending on your convention)
No, this is not true. Let us see what the correct formula is. Note that my definition of the Fourier transform is
$$\hat{f}(\xi)=\int_{\mathbb{R}^d} f(x) e^{-2\pi i x\cdot\xi} dx$$ for $\xi\in\mathbb{R}^d, f\in L^1(\mathbb{R}^d)$. This may differ from your definition by the place where you put the $2\pi$.
Assume that $u$ is a Schwartz function. Then,
$$ \begin{align} \widehat{\frac{\partial^2 u}{\partial x\partial y}}(\xi,\tau)&= \int_{\mathbb{R}^2} \frac{\partial^2 u}{\partial x\partial y}(x,y)e^{-2\pi i(x\xi+y\tau)} d(x,y)\\ &=\int_{\mathbb{R}} e^{-2\pi i y\tau} \int_{\mathbb{R}} \frac{\partial}{\partial x}\left(\frac{\partial u}{\partial y}\right)(x,y)e^{-2\pi i x\xi} dx dy\\ &=-2\pi i\xi\int_{\mathbb{R}} e^{-2\pi i y\tau} \int_{\mathbb{R}}\frac{\partial u}{\partial y}(x,y)e^{-2\pi i x\xi} dx dy \end{align} $$
where we have used Fubini's theorem in the second equality and integration by parts in the third equality.
Now use Fubini again to interchange the order of integration and use another integration by parts to get rid of the partial derivative wrt. $y$ in the same way. We get
$$\widehat{\frac{\partial^2 u}{\partial x\partial y}}(\xi,\tau)=(-2\pi i\xi)(-2\pi i \tau) \int_{\mathbb{R}}\int_{\mathbb{R}} e^{-2\pi i(x\xi+y\tau)}u(x,y)dy dx=-4\pi^2 \xi\tau \widehat{u}(x,y)$$
The same calculation can be done in higher dimensions for more partial derivatives. As you can see, the Fourier transform maps a partial derivative into multiplication with the corresponding frequency variable (up to a multiplicative constant).