$\displaystyle z=ln(3x^{2}+6y^{2}+1)$, $\displaystyle P(0,0,0)$

Ok, so I found $\displaystyle \frac{\partial f}{\partial x}$ and $\displaystyle \frac{\partial f}{\partial y}$

$\displaystyle \frac{\partial f}{\partial x}=\frac{6x}{3x^{2}+6y^{2}+1}$

$\displaystyle \frac{\partial f}{\partial y}=\frac{12y}{3x^{2}+6y^{2}+1}$

So then,

$\displaystyle f_{x}(0,0)=\frac{0}{1}=0$

$\displaystyle f_{y}(0,0)=\frac{0}{1}=0$

And I have a normal vector $\displaystyle n=(0,0,-1)$ at point $\displaystyle (0,0,0)$

So then,

$\displaystyle 0(x-0)+0(y-0)-(z-0)=0$

$\displaystyle =>z=0$

Did I do this correctly? Thanks for looking.