Inequality with directional derivative 
Suppose that $f$ is differentiable on $\mathbb{R}^2$, $\boldsymbol{l}_1$, $\boldsymbol{l}_2$ are given directions, and the intersection angle between them is $\varphi$($0<\varphi<\pi$). Prove that:
$$
\left( \frac{\partial f}{\partial x} \right) ^2+\left( \frac{\partial f}{\partial y} \right) ^2\leqslant \frac{2}{\sin ^2\varphi}\left[ \left( \frac{\partial f}{\partial \boldsymbol{l}_1} \right) ^2+\left( \frac{\partial f}{\partial \boldsymbol{l}_2} \right) ^2 \right] .
$$

Let $\boldsymbol{l}_1=(\cos \alpha,\sin \alpha)$, $\boldsymbol{l}_2=(\cos(\varphi+\alpha),\sin(\varphi+\alpha))$
$$
RHS\cdot \frac{\sin ^2\varphi}{2}=\left( \frac{\partial f}{\partial x} \right) ^2\left[ \cos ^2\alpha +\cos ^2\left( \varphi +\alpha \right) \right] +\left( \frac{\partial f}{\partial y} \right) ^2\left[ \sin ^2\alpha +\cos ^2\left( \varphi +\alpha \right) \right] +\frac{\partial f}{\partial x}\cdot \frac{\partial f}{\partial y}\left[ \sin 2\alpha +\sin \left( 2\alpha +2\varphi \right) \right] 
$$
I want to use Cauchy inequality,but $\left( \frac{\partial f}{\partial x} \right) ^2\left[ \cos ^2\alpha +\cos ^2\left( \varphi +\alpha \right) \right] +\left( \frac{\partial f}{\partial y} \right) ^2\left[ \sin ^2\alpha +\cos ^2\left( \varphi +\alpha \right) \right] \leqslant \left( \frac{\partial f}{\partial x} \right) ^4+\left( \frac{\partial f}{\partial y} \right) ^4$ is useless.
Or let $t=\frac{f_x}{f_y}$,$f(t)=\frac{\sin ^2\varphi}{2}\left( t^2+1 \right) -t\left( \cos ^2\alpha +\cos ^2\left( \varphi +\alpha \right) \right) -\frac{1}{t}\left( \sin ^2\alpha +\sin ^2\left( \varphi +\alpha \right) \right) +\sin 2\alpha +\sin \left( 2\alpha +2\varphi \right)$
$f'\left( t \right) =t\sin ^2\varphi +\frac{1}{t^2}\left( \sin ^2\alpha +\sin ^2\left( \varphi +\alpha \right) \right) -2+\sin ^2\alpha +\sin ^2\left( \varphi +\alpha \right) $,the maximum of $f(t)$ seems too hard to find.
How can I prove this inequality?

We have
$$\mathrm{RHS} - \mathrm{LHS}
= \frac{2}{\sin^2 \varphi}
\left[A\left( \frac{\partial f}{\partial x} \right) ^2 + B \left( \frac{\partial f}{\partial y} \right) ^2
+ C \frac{\partial f}{\partial x} \frac{\partial f}{\partial y} \right]
$$
where
\begin{align*}
 A &= \cos ^2\alpha +\cos ^2\left( \varphi +\alpha \right) - \frac{\sin^2 \varphi}{2}, \\
 B &= \sin ^2\alpha +\sin ^2\left( \varphi +\alpha \right) - \frac{\sin^2 \varphi}{2}, \\
 C &= \sin 2\alpha +\sin \left( 2\alpha +2\varphi \right).
\end{align*}
It suffices to prove that
$$A \ge 0, \quad B \ge 0, \quad 4AB - C^2 \ge 0.$$
For convenience, denote $c = \cos \alpha, 
~ s = \sin \alpha, ~ r = \cos \varphi, ~ t = \sin \varphi$.
We have
\begin{align*}
 A &= \frac12 t^2 + 2r^2c^2 - 2csrt ,\\
 B &= \frac12 t^2 + 2r^2s^2 + 2csrt, \\
 C &= 4c^2rt + 4cr^2s - 2rt.
\end{align*}
It is easy to prove that $A \ge 0$ and $B \ge 0$. Also, we have $4AB - C^2 = t^4 \ge 0$.
We are done.
