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I have a function $f(x,y) \in C^1$. I need to prove that if $f(x,x)=1$ for any real number $x$, then $f_x(1,1)=-f_y(1,1)$.

I tried to prove it using the definition of the partial derivative, but couldn't figure it out.

$$ f_x(1,1)=\lim\limits_{h \to 0} \frac{f(1+h,1)-f(1,1)}{h} = \lim\limits_{h \to 0} \frac{f(1+h,1)-1}{h}$$

And the same for $f_y(1,1)$.

How can I prove it?

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If $f$ is differentiable, the directional derivative along $(1,1)$ can be computed as $$ \partial_{(1,1)} f(1,1) = f'_x(1,1) \cdot 1 + f'_y(1,1)\cdot 1. $$ Since the directional derivative is zero ($f$ is constant along that direction), it must be true that $f'_x(1,1) + f'_y(1,1)=0.$

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