This is my new try: Follow the hint and I got,
$$\frac{u_{xy}}{u_x} = \frac{u_y}{u}$$
integrate both sides,
$$\ln (u_x) = \ln (u) + f(x)$$
where $f(x)$ is any function only of x. Then, we have,
$$u_x = g(x) \cdot u$$
where $g(x)$ is a function of $f(x)$ (i.e. $g(x) = e^{f(x)}$). Rewrite this equation,
$$\frac{\partial u}{\partial x} = g(x) \cdot u$$
Above is ok, but below is not: you should wright $dx$ and $du$ (because standalone $\partial u$ and $\partial x$ do not make sense), and you need to integrate rather than differentiate
$$\frac{\partial u}{u} = g(x) \partial x$$
$$-\frac{1}{u^2} = g'(x) + h(y)$$
where $h$ is any function that only of y. Therefore,
$$u = \frac{1}{\sqrt{-g'(x) - h(y)}}$$
