Multi-Variable Calc, d^2f/dxdy doesn't equal d^2f/dydx.

The Question is:
Let,
f(x,y) = xy(x^2 - y^2) / (x^2 + y^2) for (x,y) does not = (0,0)
f(x,y) = 0 for (x,y) = (0,0)
Find:
A. If f(x,y) does not = (0,0) calculate df/dx and df/dy.
B. Show that (df/dx)(0,0) = 0 = (df/dy)(0,0)
C. Show that (d^2f/dxdy)(0,0) = 1 and (d^2f/dydx)(0,0) = -1
D. What went wrong? Why are the mixed partials not equal?

***Note***
All df/dx , df/dy, d^2f/dxdy, and d^2f/dydx are partial derivatives.

A) Since f(x,y) = (x^3 y - xy^3)/(x^2 + y^2) away from the origin, the quotient rule yields
∂f/∂x = [(3x^2 y - y^3)(x^2 + y^2) - (x^3 y - xy^3) * 2x] / (x^2 + y^2)^2
∂f/∂y = [(x^3 - 3xy^2)(x^2 + y^2) - (x^3 y - xy^3) * 2y] / (x^2 + y^2)^2
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B) Use the definition of a partial derivative to answer these.
∂f(0,0)/∂x = lim(h→0) [f(0+h, 0) - f(0, 0)] / h
..............= lim(h→0) [h*0*(h^2 - 0^2)/(h^2 + 0^2) - 0] / h
..............= 0.

∂f(0,0)/∂y = lim(k→0) [f(0, 0+k) - f(0, 0)] / k
..............= lim(k→0) [0*k*(0^2 - k^2)/(0^2 + k^2) - 0] / k
..............= 0.
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C) By using parts A and B with the definition of a partial derivative,
∂²f(0,0)/∂x∂y = lim(h→0) [∂f(0+h, 0)/∂y - ∂f(0, 0)/∂y] / h
..................= lim(h→0) [(h^5 - 0)/h^4 - 0] / h
..................= 1.

∂²f(0,0)/∂y∂x = lim(k→0) [∂f(0, 0+k)/∂x - ∂f(0, 0)/∂x] / k
..................= lim(k→0) [(-k^5 - 0)/k^4 - 0] / k
..................= -1.

Hence, ∂²f(0,0)/∂x∂y ≠ ∂²f(0,0)/∂y∂x.
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D) One can check that although the first partial derivatives are continuous, the second partial derivatives are not continuous at (0, 0) (in fact, their limits as (x,y)→(0,0)) ∂²f/∂x∂y do not exist).

Link:
http://www.math.utk.edu/~denzler/M247-Sp…
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I hope this helps!