# partial derivative – Ge the canonical form of the equation

Dice $$x ^ 2u_ {xx} + 2xyu_ {xy} – 3y ^ 2u_ {yy} – 2xu_x + 4yu_y + 16x ^ 4u = 0$$
I have to find the canonical form. The determinant is the following $$D = 16x ^ 2y ^ 2$$
I considered cases when $$x$$ Y $$and$$ can be $$0$$. When calculating for the remaining case (where both $$x$$ Y $$and$$ It is not $$0$$ and the type is hyperbolic), I'm not getting its canonical form. I'm receiving this
$$(- 3y ^ 2 – 6x ^ 3y + 9x ^ 6) u _ { xi xi} + (-3y ^ 2 + frac {2y} {x} + frac {1} {x ^ 2} ) u _ {eta} + (-6y ^ 2 – 6x ^ 2 + frac {2y (1-3x ^ 4)} x) u _ { eta xi} + ( frac {-4} {x} + 4y) u_ eta + 4yu_ xi + 16x ^ 4u = 0$$
But the point is that this does not meet the requirements for the canonical form of hyperbolic type.
Any help would be appreciated.