Change of coordinates breaks the equality of mixed partial derivatives I'm trying to reduce the PDE
$$u_{xx} + 2 u_{xy} + u_{yy} = 0,$$
to its canonical form $u_{\eta \eta} = 0.$
According to the book that I'm using, $\epsilon (x,y) = y-x$ and $\eta (x,y) = x$ is a valid coordinate transformation, so lets use it.
After doing the algebra, 
$$u_{xx} = -1[u_{\epsilon \epsilon} * (-1) + u_{\epsilon \eta }] + [(-1)* u_{\eta \epsilon} + u_{\eta \eta}],$$
$$u_{yy} = u_{\epsilon \epsilon}$$
$$u_{xy} = (-1)[(-1)u_{\epsilon \epsilon}] + [u_{\eta \epsilon}],$$
but $$u_{yx} = (-1) u_{\epsilon \epsilon} + u_{\epsilon \eta} \not = u_{xy}.$$
So, what is wrong in with this transformation ?, and why the mixed partials are not equal in $\epsilon, \eta$ coordinates ? 
I mean, I plug everything into the PDE, by $u_{xy}$ does not transforms the PDE into its canonical form, but $u_{yx} $ does.
I checked my calculations, but I couldn't find an error either.
 A: It appears that you’ve got a stray factor of $-1$ in your expression for $u_{xy}$.  
We have $${\partial \over \partial x} = {\partial\epsilon \over \partial x}{\partial \over \partial\epsilon} + {\partial\eta \over \partial x}{\partial \over \partial\eta} = -{\partial \over \partial\epsilon} + {\partial \over \partial\eta} \\ {\partial \over \partial y} = {\partial\epsilon \over \partial y}{\partial \over \partial\epsilon}+{\partial\eta \over \partial y}{\partial \over \partial\eta} = {\partial \over \partial\epsilon}$$ so, assuming equality of mixed partials with respect to $\epsilon$ and $\eta$, $${\partial^2 \over \partial x \, \partial y} = \left(-{\partial \over \partial\epsilon} + {\partial \over \partial\eta}\right){\partial \over \partial\epsilon} = -{\partial^2 \over \partial\epsilon^2}+{\partial^2 \over \partial\eta\,\partial\epsilon} = -{\partial^2 \over \partial\epsilon^2}+{\partial^2 \over \partial\epsilon\,\partial\eta} = {\partial \over \partial\epsilon}\left(-{\partial \over \partial\epsilon} + {\partial \over \partial\eta}\right) = {\partial^2 \over \partial y \, \partial x}.$$
