Let $P:\mathbb R^2\to \mathbb R$ be a polynomial with no constant or linear terms, and with all other coefficients bounded by 1 in absolute value. Suppose its Hessian determinant is equal to $\epsilon Q(x,y)$ for some tiny $\epsilon$ and some other polynomial $Q$ with all coefficients bounded by 1 in absolute value. Then is it true that for some unit vector $u$, we have $$ (\nabla\cdot u) P(x,y)=\epsilon^{1/2} C_d R(x,y) $$ for some constant $C_d$ depending on the degree $d$ of $P$ only, and some polynomial $R$ with all coefficients bounded above by 1 in absolute value?
Edited: this question occurs in my research, where I need to consider the set where the Hessian determinant of a polynomial $P$ is small. There is one particular case which bothers me, namely, when $\det(D^2P)$ is of the order $O(\epsilon)$ in the entire $[0,1]^2$ but some coefficients of $P$ is large, say when $$ P(x,y)=x^2+\epsilon^{1/2}y^2. $$ In this case we see that $\partial_y P(x,y)$ is of the order $\epsilon^{1/2}$. However, as in the case $$ P(x,y)=x^2+(2+\epsilon^{1/2})xy+y^2, $$ all coefficients of $P$ are large, but taking $u=(1/\sqrt 2,1/\sqrt 2)$ also does the job. For a general polynomial with high order, this rotation may be difficult to find.
I suspect what you want is false, the example would be $$ P = x^2 + 2xy + y^2 $$ where the Hessian derivative is zero.
If this example does not finish things, you should explain why.