Answer
$v_{ss}=-sin(s^2-t^2)\times 4s^2 +2cos(s^2-t^2)$ $v_{tt}=-sin(s^2-t^2)\times4t^2-2cos(s^2-t^2)$
$v_{st}=v_{ts}=4st\times sin(s^2-t^2)$
Work Step by Step
Take the first partial derivatives of the given function. When taking partial derivative with respect to s, treat t as a constant, and vice versa:
$v_{s}=cos(s^2-t^2)\times2s$
$v_{t}=cos(s^2-t^2)\times-2t$
Then take the derivative of the first order partial derivatives to find second partial derivatives:
$v_{ss}=-sin(s^2-t^2)\times 4s^2 +2cos(s^2-t^2)$ $v_{tt}=-sin(s^2-t^2)\times4t^2-2cos(s^2-t^2)$
Second partial derivatives of first order partial derivative of s with respect to t and t with respect to s are the same:
$v_{st}=v_{ts}=4st\times sin(s^2-t^2)$