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The given differential equation is $\frac{dy}{dx} = \frac{x^2+y^2}{xy}$. This is a homogeneous differential equation because all terms in the numerator and denominator have the same degree (degree 2). For such equations, the standard substitution is $y = vx$.
Differentiate $y = vx$ with respect to $x$ using the product rule:

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

Substitute these expressions into $\frac{dy}{dx} = \frac{x^2+y^2}{xy}$:

$$v + x \frac{dv}{dx} = \frac{x^2+(vx)^2}{x(vx)}$$

$$v + x \frac{dv}{dx} = \frac{x^2+v^2x^2}{vx^2}$$

Factor out $x^2$ from the numerator:

$$v + x \frac{dv}{dx} = \frac{x^2(1+v^2)}{vx^2}$$

Cancel out $x^2$:

$$v + x \frac{dv}{dx} = \frac{1+v^2}{v}$$

Move $v$ from the left side to the right side:

$$x \frac{dv}{dx} = \frac{1+v^2}{v} - v$$

Combine the terms on the right side by finding a common denominator:

$$x \frac{dv}{dx} = \frac{1+v^2 - v^2}{v}$$ $$x \frac{dv}{dx} = \frac{1}{v}$$

Now, rearrange the equation so that all terms involving $v$ are on one side with $dv$, and all terms involving $x$ are on the other side with $dx$:

$$v \, dv = \frac{1}{x} \, dx$$

$$\int v \, dv = \int \frac{1}{x} \, dx$$

Perform the integration:

$$\frac{v^2}{2} = \ln|x| + C$$

where $C$ is the constant of integration.

Replace $v$ with $\frac{y}{x}$:

$$\frac{\left(\frac{y}{x}\right)^2}{2} = \ln|x| + C$$ $$\frac{y^2}{2x^2} = \ln|x| + C$$

Multiply both sides by $2x^2$:

$$y^2 = 2x^2(\ln|x| + C)$$

This is the general solution to the differential equation.