So, I can not find out what I'm doing wrong with this question, even if my life depended on it. I know instruction for doing it, but I can't seem to figure out what I'm doing wrong. Because I refuse to believe the normal is (-0.1292, -0.1292, 0) unless someone here confirms it. Everything seems to be running smoothly up until the creation of the tangent plane, that's where my suspicion lies at the moment.
This is the problem:
Given the function $$ f(x, y)=e^{-(x^2+y^2)} $$ a) Use $f\left(\frac{1}{2}, \frac{1}{2} \right)$ and it's partial derivatives to find a normal vector to $f(x, y)$ in the point $\left(\frac{1}{2}, \frac{1}{2} \right)$
Any help in helping me figure out is greatly appreciated. Thanks in advance.
(edit) A messy image of GeoGebra and what I've done https://i.stack.imgur.com/u67XQ.png
A function $(x,y)\mapsto f(x,y)$ has no normal, but a curve in ${\mathbb R}^2$ or a surface in ${\mathbb R}^3$ have normals. Your link indicates that you are considering the surface $$S: \quad z=f(x,y)\tag{1}$$ in ${\mathbb R}^3$ and want to know the normal of $S$ at the point $P=\bigl({1\over2},{1\over2}, f\bigl({1\over2},{1\over2}\bigr)\bigr)\in S$.
In order to find this normal we need a parametric representation of $S$. To this end we use $(x,y)$ as parameters and convert $(1)$ into $${\bf f}:\quad {\mathbb R}^2\to{\mathbb R}^3,\qquad (x,y)\mapsto\bigl(x,y,f(x,y)\bigr)\ .$$ Given the parameter point $P'=\bigl({1\over2},{1\over2}\bigr)$ we obtain $P=\bigl({1\over2},{1\over2},e^{-1/2}\bigr)$. The vectors $${\bf f}_x(P')=\bigl(1,0,f_x(P')\bigr),\quad {\bf f}_y(P')=\bigl(0,1,f_y(P')\bigr)$$ are (linearly independent) tangent vectors to $S$ at $P$. Therefore the vector $${\bf n}:={\bf f}_x(P')\times{\bf f}_y(P')=\bigl(-f_x(P'),\>-f_y(P'),\>1\bigr)$$ is a normal vector to $S$ at $P$.