Find all $x,y$ for which $\nabla f$ forms an angle of $45°$ with the vector $(1,1)$ Let $f=x^2+y^2 \implies \nabla f=(2x,2y)$.
Find all $x,y$ for which $\nabla f$ forms an angle of $45°$ with the vector $(1,1)$.
So I thought of taking the dot product $2x+2y=\nabla f \cdot v=||\nabla f|\cdot||v|| \cos(45°)=4\sqrt{x^2+y^2}$.
But the I couldn't solve $x+y=2\sqrt{x^2+y^2}$, I know, from inspection that $x=-y$ is a solution, and I think there are no other solutions, but I don't know how to prove it.
Could you guys help me out?
E: Oh god, I see where my mistake was... I miscalculated $\cos(45°)$ as $\sqrt 2$.

I'll do the same thing as you, but I think that that your dot product is incorrect. If $v$ is the vector through $(1,1)$, then $v=(\frac{1}{2},\frac{1}{2})$, so
$$
x+y=\nabla f \cdot v = ||\nabla f|| \cdot ||v|| \cos{(45^{o})}= \sqrt{x^2+y^2}.
$$
Square that and obtain
$$
(x+y)^2=x^2+y^2 \implies 2xy=0 \implies xy=0,
$$
so $x=0$ or $y=0$, which means that $(x,y)=(a,0)$ or $(x,y)=(0,b)$, $a,b \in \mathbb{R}^{+}.$
