Q.$\left(x^3+y^3\right)dy-x^2y\,dx=0$

A function $f(x,y)$ is a homogeneous function of degree $n$ if $f(tx,ty)=t^n f(x,y)$ for every suitably restricted $x,y,t$ (Euler's homogeneity). A homogeneous function of degree zero can always be written purely as a function of the single ratio $\dfrac{y}{x}$ (or $\dfrac{x}{y}$): $f(x,y)=g\!\left(\dfrac{y}{x}\right)$.

Homogeneous differential equation. An ODE is in homogeneous form if it can be written as

Equivalently, $M(x,y)\,dx+N(x,y)\,dy=0$ is homogeneous exactly when $M$ and $N$ are homogeneous functions of the same degree — because then $f(x,y)=-M/N$ is automatically homogeneous of degree $0$. (This use of the word "homogeneous" for the equation is a different meaning from calling the constant term $g(x)=0$ in a linear equation "homogeneous" — Definition 10.7 versus Definition 10.12 in the textbook — so the two uses should not be confused.)

Solution method (Theorem 10.1). Substitute $y=vx$ (so $v=\dfrac{y}{x}$), giving $\dfrac{dy}{dx}=v+x\dfrac{dv}{dx}$. The homogeneous equation becomes

which is variables-separable in $v$ and $x$:

Integrate both sides, then replace $v$ by $\dfrac{y}{x}$ to return to the original variables.

When to substitute $x=vy$ instead. If the equation naturally involves the ratio $\dfrac{x}{y}$ (for instance $\dfrac{dx}{dy}=g\!\left(\dfrac{x}{y}\right)$, or the algebra is simply cleaner that way), put $x=vy$ instead, giving $\dfrac{dx}{dy}=v+y\dfrac{dv}{dy}$, and proceed by the same separation-of-variables logic with the roles of $x$ and $y$ exchanged.

Practical check. To test whether a given equation is homogeneous, rewrite $\dfrac{dy}{dx}$ as a ratio of two expressions in $x,y$ and confirm that scaling both $x\to tx,\ y\to ty$ leaves the ratio unchanged (the $t$'s cancel) — equivalently, that every term in $M$ and $N$ has the same total degree in $x$ and $y$.