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,

$$u = \frac{1}{\sqrt{-g'(x) - h(y)}}$$