Answer
$dz/dt=( y^3-2xy)(2t)+(3xy^2-x^2)(2t)$
or
$dz/dt=( (t^2-1)^3-2(t^2+1)(t^2-1))(2t)+(3(t^2+1)(t^2-1)^2-(t^2+1)^2(2t))$
Work Step by Step
Chain Rule is:
$dz/dt=(z_{x})(dx/dt)+(z_{y})(dy/dt)$
Find partial derivatives for z with respect to x and y:
$z_{x} = y^3-2xy$
$z_{y}=3xy^2-x^2$
Find the first order derivatives of x and y:
$dx/dt=2t$
$dy/dt=2t$
Plug in the solved parts into the Chain rule to find dz/dt:
$dz/dt=( y^3-2xy)(2t)+(3xy^2-x^2)(2t)$
Plug in the x and y equations into dz/dt to make all variables t:
$dz/dt=( (t^2-1)^3-2(t^2+1)(t^2-1))(2t)+(3(t^2+1)(t^2-1)^2-(t^2+1)^2(2t))$
