Let $p\in \mathbb{R}[x,y]$ be a homogeneous polynomial of odd degree $d \geq 1$.
Can one find an affine $\mathbb{R}$-algebra automorphism $G$ of $\mathbb{R}[x,y]$ such that $G(p)_x(\mathbb{R}^2) \geq 0$, where $G(p)_x$ is the partial derivative of $G(p) \in \mathbb{R}[x,y]$ with respect to $x$? Notice that $G(p)_x$ is homogeneous of even degree $d-1$.
I do not mind to further assume that the coefficients of $p$ belong to $\mathbb{R}^{\geq 0}=[0,\infty)$, if this simplifies the answer.
Examples:
(1) $d=1$: Write $p=ax+by$, $a,b \in \mathbb{R}$.
If $a \geq 0$ then for $G=1$ we have $G(p)=p=ax+by$, so $G(p)_x=a \geq 0$. If $a < 0$ then for $G: (x,y) \mapsto (-x,y)$ we have $G(p)=G(ax+by)=aG(x)+bG(y)=a(-x)+by=-ax+by$, so $G(p)_x=-a > 0$.
(2) $d=3$: Write $p=ax^3+bx^2y+cxy^2+dy^3$, $a,b,c,d \in \mathbb{R}$.
If $a \neq 0$, then we can define $G: (x,y) \mapsto (x-\frac{b}{3a}y,y)$. Then a direct computation (if I do not have an error) yields $G(p)=ax^3+(\frac{-b^2}{3a}+c)xy^2+\epsilon y^3$, for some $\epsilon \in \mathbb{R}$. If we further assume that $a,b,c,d \in \mathbb{R}^{+}$, then we are almost done; indeed, $G(p)_x=3ax^2+(\frac{-b^2}{3a}+c)y^2$, which is non-negative in case $\frac{-b^2}{3a}+c \geq 0$, so we need $3ac-b^2 \geq 0$.
(Allowing $G$ to be an affine $\mathbb{C}$-algebra automorphism of $\mathbb{C}[x,y]$ still does not help; if we take $H: (x,y) \mapsto (x-\frac{bi}{3a}y,y)$ instead of our former $G$, then $\delta xy$ will appear in $H(p)_x$, for some $\delta \in \mathbb{C}$).
(3) $d=5$: Write $p=ax^5+bx^4y+cx^3y^2+dx^2y^3+exy^4+fy^5$, $a,b,c,d,e,f \in \mathbb{R}^{+}$. "Completing the square" trick, as far as I see, only guarantees that we can find an affine $G$ such that $G(p)=Ax^5+Cx^3y^2+Dx^2y^3+Exy^4+Fy^5$ for some $A,C,D,E,F \in \mathbb{R}$, but this is not enough, because $G(p)_x=5Ax^4+3Cx^2y^2+2Dxy^3+Ey^4$, which contains the problematic term $2Dxy^3$. Actually, one also has to be careful about the signs of $A,C,E$, which are dependent on the relations between the original coefficients $\{a,b,c,d,e,f\}$ (even if all of them are positive), as we have seen in the case $d=3$.
Any comments are welcome!
See also this question.