How do we find the critical point using partial derivatives?

If $\frac{\partial f}{\partial x}= 0$ when $x = 1$, and $\frac{\partial f}{\partial x} = (2y-x)$

Do we have to replace $x$ by 1? to get the critical coordinate $y$?

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Indeed you have :) –  Pragabhava Dec 6 '12 at 0:20
The critical points occur when $f_{x}=f_{y}=0$. So, necessarily, any critical point must occur when $x=1$ so that we obtain $2y-1=0$ and $y=\frac{1}{2}$ as desired. So you are correct.
Note that just because $(1, \frac{1}{2})$ is a potential critical point doesn't mean it is one--you still need to check $f_y$ –  anorton Dec 6 '12 at 0:34
I disagree... My textbook ("University Calculus" by Hass, Weir, and Thomas) defines a critical point as "An interior point of the domain of a function $f(x,y)$ where both $f_x$ and $f_y$ are zero or where one or both of $f_x$ and $f_y$ do not exist is a critical point of $f$." Wikipedia also verifies this: en.wikipedia.org/wiki/… –  anorton Dec 6 '12 at 1:26
Oh. Sorry. I meant it has the potential to be a critical point. It could be, but we don't know yet. To verify, we must know $f_y$. –  anorton Dec 6 '12 at 1:29