Local Minimum and Maximum; Saddle Point

Hi everyone. I have a question on one of my homework problems.

Q. Find the local max./min. values and the saddle point(s) of the function $\displaystyle f(x,y) = (x^2+y^2)e^{y^2-x^2}$

So I started doing this problem and it was looking really messy so I went on cramster and got this:

Step 1

http://s3.amazonaws.com/answer-board...4062504777.gif

Step 2

http://s3.amazonaws.com/answer-board...1250005773.gif

http://s3.amazonaws.com/answer-board...0000007060.gif

http://s3.amazonaws.com/answer-board...9687502814.gif

http://s3.amazonaws.com/answer-board...4062501171.gif

Step 3

http://s3.amazonaws.com/answer-board...0000006297.gif

http://s3.amazonaws.com/answer-board...3437504755.gif

http://s3.amazonaws.com/answer-board...2812502914.gif

http://s3.amazonaws.com/answer-board...9062507561.gif

so critical points are (0,0), (1,0), and (-1,0)

Step 4

http://s3.amazonaws.com/answer-board...2812505189.gif

http://s3.amazonaws.com/answer-board...4843750803.gif

http://s3.amazonaws.com/answer-board...0625002996.gif

Step 5

http://s3.amazonaws.com/answer-board...1875005247.gif

at (0,0): http://s3.amazonaws.com/answer-board...5625009888.gif

at (1,0): http://s3.amazonaws.com/answer-board...9062506345.gif

at (-1,0): http://s3.amazonaws.com/answer-board...5000007320.gif

Therefore, f has a local minimum of 0 at (0,0) and saddle points at (1,0) and (-1,0).

The answer is correct but the work makes no sense. When computing $\displaystyle f_x$, $\displaystyle y^2e^{y^2-x^2}$ was completely neglected. $\displaystyle f_x$ should have been:

$\displaystyle f_x = 2xe^{y^2-x^2}-2x^3e^{y^2-x^2} -2xy^2e^{y^2-x^2} = 0$

Same problem with $\displaystyle f_y$

Is there something wrong with the solution or am I doing something wrong?

Thanks in advance!