Let us consider a $C^1$ function $f : \mathbb R_+^* \times \mathbb R_+^* \to \mathbb R_+^*$.
If $f$ is homogeneous of degree $d$, then $\forall (x,y) \in \mathbb R_+^* \times \mathbb R_+^*, \frac{\partial_x f(\lambda x, \lambda y)}{\partial_y f(\lambda x, \lambda y)}$ is independent of $\lambda \in \mathbb R_+^*$.
What about the reciprocal?
If we have that $\forall (x,y) \in \mathbb R_+^* \times \mathbb R_+^*, \frac{\partial_x f(\lambda x, \lambda y)}{\partial_y f(\lambda x, \lambda y)}$ does not depend on $\lambda \in \mathbb R_+^*$, does that mean $f$ is homogeneous for some $d$ (up to an additive constant)?
Take $f(x,y)=\exp(xy)$. It is not homogeneous, but satisfies the weaker condition $$ \frac{\partial_x f(\lambda x,\lambda y)}{\partial_y f(\lambda x,\lambda y)}= \frac yx $$ independent of $\lambda$. Actually, any $f(x,y)=g(x^ay^b)$ would do, for an arbitrary function $g$ of a single variable.
Even just $f(x,y)=g(y)$ for an arbitrary function $g$ of single variable satisfies the weaker condition, irrespectively of the whether $g$ is homogeneous or not.