Find the solution $u$ of $x~u~u_x+y~u~u_y=-x~y,$ that $u(x,1/x)=5,~x>0$. 
Solve the following Cauchy problem:
$$x~u~u_x+y~u~u_y=-x~y,$$ under the condition that $u(x,1/x)=5,~x>0$.

Attempt.
The characteristic curves for this quasilinear pde satisfy the system of equations:
$$\frac{dx}{xz}=\frac{dy}{yz}=\frac{dz}{-xy}.$$
From $\displaystyle \frac{dx}{xz}=\frac{dy}{yz}$, we easily get $g_1(x,y,z)=y/x=c_1$. We are still looking for a second surface $g_2(x,y,z)=c_2$, such that $\nabla g_1\times \nabla g_2\neq (0,0,0)$, so that the general solution can be given as $F(g_1,g_2)=0,$ for $F\in C^1.$ This is where I am stuck: so far I have not figured out a standard method of solving systems, like the above - this is a case where one integration comes easily, while the rest do not seem to me that obvious.
Thank you in advance.

$$x~u~u_x+y~u~u_y=-x~y,$$ under the condition that $u(x,1/x)=5,~x>0$.
You attempt is correct, but mut be completed :
$$\frac{dx}{xu}=\frac{dy}{yu}=\frac{du}{-xy}.$$
The equation of a first characteristic curve comes from $\frac{dx}{xu}=\frac{dy}{yu}$
$$\frac{y}{x}=c_1$$
The equation of a second characteristic curve comes from :
$\frac{dx}{xu}=\frac{dy}{yu}=\frac{ydx+xdy}{y(xu)+x(yu)}=\frac{d(xy)}{2xyu}=\frac{du}{-xy} \quad\to\quad d(xy)=-2udu$
$$u^2+xy=c_2$$
The general solution expressed on the form of implicit equation is :
$$\Phi\left(\frac{y}{x}\;,\;u^2+xy\right)=0$$
where $\Phi$ is any differentiable function of two variables.
Solving the implicit equation for the second variable leads to the explicit form of general solution :
$$u^2=-xy+F\left(\frac{y}{x}\right)$$
where $F$ is any differentiable function.
The condition $u(x,1/x)=5$ implies $u^2(x,1/x)=25=-x\frac{1}{x}+F\left(\frac{\frac{1}{x}}{x}\right) \quad\to\quad F\left(\frac{1}{x^2}\right)=26$
Thus, $F$ is a constant function equal to 26.
$$u^2(x,y)=-xy+26$$
