1. Given differential equation

$x\frac{dy}{dx}=y\left(\log y-\log x+1\right)$

We rewrite it as

$\frac{dy}{dx}=\frac{y}{x}\left(\log \frac{y}{x}+1\right).$

2. Use substitution

Since the expression contains $\log\left(\frac{y}{x}\right)$, let

$v=\frac{y}{x} \quad \Rightarrow \quad y=vx.$

Then

$\frac{dy}{dx}=v+x\frac{dv}{dx}.$

Substitute into the differential equation:

$x\left(v+x\frac{dv}{dx}\right)=vx\left(\log v+1\right).$

Divide by $x$:

$v+x\frac{dv}{dx}=v(\log v+1).$

So,

$x\frac{dv}{dx}=v\log v.$

3. Separate variables

$\frac{dv}{v\log v}=\frac{dx}{x}.$

Integrate both sides:

$\int \frac{dv}{v\log v}=\int \frac{dx}{x}.$

Now,

so we get

$\log|\log v|=\log|x|+C.$

This implies

$\log v=Cx$

(after absorbing constants appropriately).

Since $v=\frac{y}{x}$,

$\log\left(\frac{y}{x}\right)=Cx.$

4. Match with options

This is exactly:

$\boxed{\log\left(\frac{y}{x}\right)=cx}$

So the correct option is C.