Find $d^3z$ of $z = x\cos(y)+y\sin(x)$
My Attempt
$$dz = \frac{\partial z}{\partial x}dx + \frac{\partial z}{\partial y}dy = (\cos(y)+y\cos(x))dx + (-x\sin(y) + \sin(x))dy$$
$$d^2z = (-\sin(y)dy+\cos(x)dy - y\sin(x)dx)dx + (-\sin(y)dx -x\cos(y)dy + \cos(x)dx)dy $$ $$ = - y\sin(x)dx^2 + (-2\sin(y)+ 2\cos(x))dxdy -x\cos(y)dy^2 $$
$$ d^3z = (- \sin(x)dy -y\cos(x)dx)dx^2 + (-2\cos(y)dy - 2\sin(x)dx)dxdy+( -\cos(y)dx + x\sin(y)dy)dy^2 $$
$$ d^3z = - \sin(x)dydx^2 -y\cos(x)dx^3 + -2\cos(y)dxdy^2 - 2\sin(x)dx^2dy) -\cos(y)dxdy^2 + x\sin(y)dy^3 $$
$$ d^3z = -y\cos(x)dx^3 + -3\cos(y)dxdy^2 - 3\sin(x)dx^2dy)+ x\sin(y)dy^3 $$
Is this the correct approach / answer?