The given differential equation is: \( xe^{\frac{y}{x}} - y + x\frac{dy}{dx} = 0 \)

Rearrange the equation to isolate \(\frac{dy}{dx}\):

\( x\frac{dy}{dx} = y - xe^{\frac{y}{x}} \)

\( \frac{dy}{dx} = \frac{y}{x} - e^{\frac{y}{x}} \)

This is a homogeneous differential equation. Let \( y = vx \), so \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)

Substitute \( y = vx \) into the equation:

\( v + x\frac{dv}{dx} = v - e^v \)

Simplify the equation:

\( x\frac{dv}{dx} = -e^v \)

Separate the variables:

\( e^{-v} dv = -\frac{dx}{x} \)

Integrate both sides:

\( \int e^{-v} dv = \int -\frac{dx}{x} \)

\( -e^{-v} = -\ln|x| + C \)

Multiply by -1:

\( e^{-v} = \ln|x| - C \)

Substitute back \( v = \frac{y}{x} \):

\( e^{-\frac{y}{x}} = \ln|x| - C \)

We can rewrite the constant as \( \ln|C_1| \) for simplicity:

\( e^{-\frac{y}{x}} = \ln|x| + \ln|C_1'| \)

\( e^{-\frac{y}{x}} = \ln|C_1 x| \)

Taking exponential on both sides is not necessary, the solution is:

\( e^{-\frac{y}{x}} = \ln|x| + C \)