The given differential equation is:

$$y \log y \, dx - x \, dy = 0$$

To find the general solution, we first rearrange the terms to separate the variables $x$ and $y$. Moving $x \, dy$ to the right side, we get:

$$y \log y \, dx = x \, dy$$

Next, we separate the variables by dividing both sides by $xy$. This gives us the equation:

$$\frac{dy}{y \log y} = \frac{dx}{x}$$

We assume that $x \neq 0$ and $y \neq 0$ (and $\log y$ exists, so $y > 0$).

Now, we integrate both sides of the equation with respect to their respective variables. The equation becomes:

$$\int \frac{dy}{y \log y} = \int \frac{dx}{x}$$

For the left-hand side, we use the substitution method: let $t = \log y$, which implies $dt = \frac{1}{y} dy$. The integral becomes:

$$\int \frac{1}{t} \, dt$$

Evaluating the integrals, we get:

$$\log |t| = \log |x| + C_1$$

Substituting $t = \log y$ back into the equation, we have:

$$\log |\log y| = \log |x| + \log C$$

(where we rewrite the constant $C_1$ as $\log C$ for convenience).

Finally, using the property $\log A + \log B = \log AB$, we simplify the equation to:

$$\log |\log y| = \log |Cx|$$

Taking the antilogarithm, we get:

$$\log y = Cx$$

(where $C$ is an arbitrary constant). This is the required general solution.