I want to find all unit vectors $\textbf{v}$ such that the directional derivative of a function $f(x,y)=x^2y^2$ in the direction $\textbf{v}$ at $(1,\sqrt{3})$ is half the value of the directional derivative in the direction $\textbf{u}$ where $f$ increases most rapidly at the same point.
Here, we have $\nabla f(1,\sqrt{3})=(6,2\sqrt{3})$ and $|\nabla f(1,\sqrt{3})|=4\sqrt{3}$. Thus $$ \textbf{u}=\frac{\nabla f(1,\sqrt{3})}{|\nabla f(1,\sqrt{3})|}=\left(\frac{\sqrt{3}}{2},\frac{1}{2}\right) $$ and $D_{\textbf{u}}f(1,\sqrt{3})=4\sqrt{3}$.
My approach: We have $$ D_{\textbf{v}}f(1,\sqrt{3})=\frac12D_{\textbf{u}}f(1,\sqrt{3}), $$ and so, with $\textbf{v}=(a,b)$, $$ (a,b)\cdot (6,2\sqrt{3})=\frac12|\nabla f(1,\sqrt{3})|=2\sqrt{3}\Rightarrow a=\frac{1-b}{\sqrt{3}}. $$ Now, since we need $|\textbf{v}|=1$, we further impose $$ \left(\frac{1-b}{\sqrt{3}} \right)^2+b^2=1\Rightarrow b=1\text{ or }b=-\frac12. $$ Hence, we have two solutions, $\textbf{v}_1= \left(0 , 1 \right)$ and $\textbf{v}_2= \left( \frac{\sqrt{3}}{2} ,-\frac12 \right)$.
Is this correct? I could multiply by a rotation matrix, knowing the angle, but I wanted to avoid that. Is there a quicker way?