Here's another method in which we avoid any risk of accidentally dividing by $0$, I'm assuming here that $x,y$ are not dependent--that is, neither is a function with the other as a variable, and so $\frac{\partial x}{\partial y}=\frac{\partial y}{\partial x}=0$. Let me know if that's incorrect, and I'll adapt my answer accordingly.
We may first observe that $r^2=x^2+y^2$, so since $\frac{\partial x}{\partial y}=0$ and $y=r\sin\theta$, then $2r\frac{\partial r}{\partial y}=2x\frac{\partial x}{\partial y}+2y=2y=2r\sin\theta.$ Now, there are two ways this can happen: either (i) $r=0$ (in which case your given expression for the mixed partial cannot hold) or (ii) $r\neq 0$ and $\frac{\partial r}{\partial y}=\sin\theta$. Similarly, either (i) $r=0$ or (ii) $r\neq 0$ and $\frac{\partial r}{\partial x}=\cos\theta$.
Let's see what happens in the case that $r\neq 0$, $\frac{\partial r}{\partial y}=\sin\theta$, and $\frac{\partial r}{\partial x}=\cos\theta$. Since $x=r\cos\theta$, $\frac{\partial x}{\partial y}=0$, and $\frac{\partial r}{\partial y}=\sin\theta$, then by product and chain rules, we have $0=\cos\theta\frac{\partial r}{\partial y}-r\sin\theta\frac{\partial\theta}{\partial y}=\cos\theta\sin\theta-r\sin\theta\frac{\partial\theta}{\partial y}.\tag{1}$ Similarly, since $y=r\sin\theta$ and $\frac{\partial r}{\partial y}=\sin\theta$, then $1=\sin^2\theta+r\cos\theta\frac{\partial\theta}{\partial y}.\tag{2}$ With some rearrangement, dividing through by $r\neq 0$, and an application of the Pythagorean Identity, we have $\frac1r\cos\theta\sin\theta=\sin\theta\frac{\partial\theta}{\partial y},\tag{$1'$}$ $\frac1r\cos^2\theta=\cos\theta\frac{\partial\theta}{\partial y}.\tag{$2'$}$ Now, since $\frac{\partial r}{\partial x}=\cos\theta$, then differentiating $(2')$ with respect to $x$, we have again by chain and product rules, and by substitution using $(1')$, that
$\begin{eqnarray*} -2\cos\theta\sin\theta\frac{\partial\theta}{\partial x} & = & \cos\theta\frac{\partial\theta}{\partial y}\frac{\partial r}{\partial x}-r\sin\theta\frac{\partial\theta}{\partial y}\frac{\partial\theta}{\partial x}+r\cos\theta\frac{\partial^2\theta}{\partial x\partial y}\\ & = & \cos^2\theta\frac{\partial\theta}{\partial y}-r\sin\theta\frac{\partial\theta}{\partial y}\frac{\partial\theta}{\partial x}+r\cos\theta\frac{\partial^2\theta}{\partial x\partial y}\\ & = & \cos^2\theta\frac{\partial\theta}{\partial y}-\cos\theta\sin\theta\frac{\partial\theta}{\partial x}+r\cos\theta\frac{\partial^2\theta}{\partial x\partial y}, \end{eqnarray*}$
and so $0=\left(\cos\theta\frac{\partial\theta}{\partial y}+\sin\theta\frac{\partial\theta}{\partial x}+r\frac{\partial^2\theta}{\partial x\partial y}\right)\cos\theta.\tag{3}$
Since $\frac{\partial r}{\partial x}=\cos\theta$, then differentiating $(1')$ with respect to $x$, we have again by chain and product rules, and by substitution using $(2')$, that
$\begin{eqnarray*} (\cos^2\theta-\sin^2\theta)\frac{\partial\theta}{\partial x} & = & \sin\theta\frac{\partial\theta}{\partial y}\frac{\partial r}{\partial x}+r\cos\theta\frac{\partial\theta}{\partial y}\frac{\partial\theta}{\partial x}+r\sin\theta\frac{\partial^2\theta}{\partial x\partial y}\\ & = & \cos\theta\sin\theta\frac{\partial\theta}{\partial y}+r\cos\theta\frac{\partial\theta}{\partial y}\frac{\partial\theta}{\partial x}+r\sin\theta\frac{\partial^2\theta}{\partial x\partial y}\\ & = & \cos\theta\sin\theta\frac{\partial\theta}{\partial y}+\cos^2\theta\frac{\partial\theta}{\partial x}+r\sin\theta\frac{\partial^2\theta}{\partial x\partial y}, \end{eqnarray*}$
and so $0=\left(\cos\theta\frac{\partial\theta}{\partial y}+\sin\theta\frac{\partial\theta}{\partial x}+r\frac{\partial^2\theta}{\partial x\partial y}\right)\sin\theta.\tag{4}$
Now, $\cos\theta$ and $\sin\theta$ are never simultaneously $0$, so by $(3)$ and $(4)$, it follows that $\cos\theta\frac{\partial\theta}{\partial y}+\sin\theta\frac{\partial\theta}{\partial x}+r\frac{\partial^2\theta}{\partial x\partial y}=0$ in any case, so since $r\neq 0$, then $\frac{\partial^2\theta}{\partial x\partial y}=-\frac1r\cos\theta\frac{\partial\theta}{\partial y}-\frac1r\sin\theta\frac{\partial\theta}{\partial x},$ whence substitution using $(2')$ yields $\frac{\partial^2\theta}{\partial x\partial y}=-\frac{\cos^2\theta}{r^2}-\frac1r\sin\theta\frac{\partial\theta}{\partial x}.\tag{5}$ Since $x=r\cos\theta$ and $\frac{\partial r}{\partial x}=\cos\theta$, then in a similar fashion to our development of $(2')$ we find that $\frac1r\sin^2\theta=-\sin\theta\frac{\partial\theta}{\partial x},\tag{6}$ and subbing this into $(5)$ gives us $\frac{\partial^2\theta}{\partial x\partial y}=-\frac{\cos^2\theta-\sin^2\theta}{r^2}.\tag{$5'$}$ Finally, a double angle identity for cosine gives us the desired result.