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Could one have a function $f(x,y)$ s.t. it is increasing along the line $x=y$ but the partial derivatives $\frac{\partial f(x,y)}{\partial y} = \frac{\partial f(x,y)}{\partial x} = 0$ on every point on that line.

Essentially this would amount to every point on that line being a saddle point.

The function is differentiable everywhere.

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No, because the directional derivative in the direction of $\vec{v}=(1,1)$ is nonzero along $x=y$, but the directional derivative is a linear combination of partial derivatives. If $\vec{u}=\vec{v}/|\vec{v}|$, then $D_{\vec{u}}f=\nabla f\cdot \vec u$, and your condition is that $\nabla f$ vanishes along the line $x=y$, which means that all directional derivatives are also zero, so the function can't be monotone increasing in any direction.

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