Partial derivative stationary point

[Taylor_1989]

Homework Statement

Hi guys, I am having real trouble with the function 10ii) I can take the derivatives, but I feel like I am missing something, with what I have done. I set $f_x=0$and $f_y=0$ but really can't seem to find away to solve, i keep getting (0,0) which when I plug into wolfram it come out with $$x=0,y=-1/2$$, $$x=-1,y=0$$ and $$x=1, y=-2$$
So this make me think have I done the partial right?

Homework Equations

The Attempt at a Solution

[/B]
$$\frac{\partial}{\partial x}=-2(x^2y-x-1)(2xy-1)-2(x^2-1)(2x)$$
$$\frac{\partial}{\partial y}=-2(x^2y-x-1)(x^2)$$$$-2(x^2y-x-1)(2xy-1)-2(x^2-1)(2x)=0$$
$$-2(x^2y-x-1)(x^2)=0$$

but where do I go from here?

Could someone please give advice, thanks in advance.

 

[BvU]

I set $f_x=0\$ and $f_y=0$

Two hashes (#) to get inline $\TeX$, two dollars ($) to get displayed math

Check $f_x$ and $f_y$ (*); once OK, $f_y$ gives you something to substitute in $f_x=0$

(*) [edit] right, thanks Dick (I copied the ² inside the [ ] )

 

[Dick]

The partials look ok. Think about the y derivative first. It's zero if i) x=0 OR y=?. Fill in the ? with an expression in x. Substitute that into the x derivative.

 

[epenguin]

Not every function of x and y has a stationary point - it is always possible that your function has none.
In that case you will not be able to find it! But then you have to prove it doesn't exist.
In this case I think there is a stationary point.
f_y is nicely factorised, which is help. One of the factors is closely related to a bracket in f_x which is another help.

Not every condition for one of your derivatives to be 0 necessarily allows the the other to be 0.
(Though even if it doesn't, your zeros of one of them still gives information about what the surface f looks like)