Partial derivative problem on absolute value function Find the first and second order partial derivatives of 
$f(x,y)=|2x^2-y|$.
I start with limit definition but not able to solve.
please help.

We define $g$ as the function inside the absolute value. (i.e: $f(x,y) = |g(x,y)|$)
$$\frac{\partial f(x,y)}{\partial x} = \frac{g(x,y)\frac{\partial g(x,y)}{\partial x}}{|g(x,y)|}$$
Therefore, applying this to our function (i.e: $g(x,y) = 2x^2-y$) we get:
$$\frac{\partial |2x^2-y|}{\partial x} = \frac{(2x^2-y)(4x)}{|2x^2-y|},$$
$$\frac{\partial |2x^2-y|}{\partial y} = \frac{-2x^2+y}{|2x^2-y|},$$
for the first partial derivatives and
$$\frac{\partial\frac{(2x^2-y)(4x)}{|2x^2-y|}}{\partial x} = \frac{|2x^2-y|(24x^2-4y) - \frac{8x^3-4xy}{|2x^2-y|}}{|2x^2-y|^2},$$
$$\frac{\partial \frac{-2x^2+y}{|2x^2-y|}}{\partial y} = \frac{|2x^2-y| - \frac{(-2x^2+y)^2}{|2x^2-y|}}{|2x^2-y|^2},$$
for the second partial derivatives.
Finally, if we apply the definition of absolute value function to our results we get exactly what Statish Ramanathan said.
