Can anyone check my solution for this exercise?
Let $F:\mathbb{R}^3\rightarrow\mathbb{R}$ be given by $F(x,y,z) = \alpha xz + x\arctan(z) + z\sin(2x+y) -1.$
Prove that a function $z=f(x,y)$ can be defined around $(0, \pi/2,1)$ and find $\alpha$ such that $(0, \pi/2)$ is a critical point of $f$.
To show that $z=f(x,y)$ can be defined around $(0,\pi/2,1)$ I should apply the implicit function theorem, this is, I should check that $F(0,\pi/2,1)=0$ and $\displaystyle\frac{\partial F}{\partial z}(0,\pi/2,1) \neq .0$
I have $F(0,\pi/2,1) = \alpha(0)(1) + (0)\arctan(1) + (1)\sin(\pi/2) -1 = 0$; and also $\displaystyle\frac{\partial F}{\partial z} = \alpha x + \displaystyle\frac{x}{1+z^2}+\sin(2x+y)$ which means that $\displaystyle\frac{\partial F}{\partial z}(0,\pi/2,1) = \sin(\pi/2) = 1 \neq 0$. Then by the implicit function theorem a function $z=f(x,y)$ can be defined.
Now to find $\alpha$ such that $(0,\pi/2)$ is a critical point, this means that must be $\displaystyle\frac{\partial f}{\partial x} = \displaystyle\frac{\partial f}{\partial y} = 0$.
Considering that $\displaystyle\frac{\partial F}{\partial x} = \alpha z + \arctan z + 2z\cos(2x+y), \displaystyle\frac{\partial F}{\partial y} = z\cos(2x+y)$ and
$\displaystyle\frac{\partial f}{\partial x} = \displaystyle\frac{\frac{\partial F}{\partial x} (x,y,f(x,y))}{\frac{\partial F}{\partial z}(x,y,f(x,y))}$, $\displaystyle\frac{\partial f}{\partial y} = \displaystyle\frac{\frac{\partial F}{\partial y} (x,y,f(x,y))}{\frac{\partial F}{\partial z}(x,y,f(x,y))}$.
If $(x,y) = (0,\pi/2)$ then $F(x,y,z) = 0$ implies that $z=1$. Also $\displaystyle\frac{\partial F}{\partial x}(0,\pi/2,1) = \alpha + \arctan(1), \displaystyle\frac{\partial F}{\partial y}(0,\pi/2,1) = 0$ and $\displaystyle\frac{\partial F}{\partial z}(0,\pi/2,1) = 1$ .
Thus, $\displaystyle\frac{\partial f}{\partial y} = 0$ and $\displaystyle\frac{\partial f}{\partial x} = \alpha + \arctan(1) \implies \alpha = -1$
Am I missing something?.
Your reasoning looks pretty good to me. Do you have a reason for believing it might be wrong?