# Proportional second derivatives imply proportional first

Let $$f:[0,\infty)^n\to[0,\infty),\ f\in\mathcal{C}^{\infty}$$, if I assume that the following relation holds: $${\partial^2 f\over \partial x_1^2}\bigg\vert_{x_1=x_2=\dots=x_n=t}\propto{\partial^2 f\over \partial x_2^2}\bigg\vert_{x_1=x_2=\dots=x_n=t}\propto\cdots\propto{\partial^2 f\over \partial x_n^2}\bigg\vert_{x_1=x_2=\dots=x_n=t},\ t\in[0,\infty).$$ Does this imply that: $${\partial f\over \partial x_1}\bigg\vert_{x_1=x_2=\dots=x_n=t}\propto{\partial f\over \partial x_2}\bigg\vert_{x_1=x_2=\dots=x_n=t}\propto\cdots\propto{\partial f\over \partial x_n}\bigg\vert_{x_1=x_2=\dots=x_n=t},\ t\in[0,\infty)$$ where $$\propto$$ meaning that the functions are equal up to a constant. Any help is appreciated.

Let $$f:[0,+\infty)^2\to [0+\infty):(x,y)\mapsto x^2+x+y^2$$.
Then for all $$t\ge 0:$$ $$\partial_{11}f(t,t)=2=\partial_{22}f(t,t),$$ but $$\partial_1f(t,t) = 2t+1\\ \partial_2f(t,t)=2t.$$