If $f(x) = |x{|^3},$ show that f$''(x)$ exists for all real x and find it

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Case I. When $x \ge 0.$
Here,$f(x) = |x{|^3} = {x^3}$

therefore, $f'(x) = 3{x^2}$ and $f''(x) = 6x$

Case II. When x $<$ 0.
Here $f(x) = {( - x)^3} = - {x^3}$

therefore, $f'(x) = - 3{x^2}$ and $f''(x) = -6x$

Hence, we can say that f$''(x)$ exist for all real x.