EDIT: I read the question wrong, I need an efficient way to calculate $f_{yyzzx}$:EDIT
Consider the function $f(x,y,z,w)=x^2\sin(4y)+z^3(6x-y)+y^4$. Use Clairut's Theorem to find $f_{yyzzx}$
$f_y=4x^2\cos(4y)-z^3+4y^3$
$f_{yy}=-16x^2\sin(4y)+12y^2$
For $f_{yyz}$ I'm getting confused, because if I'm doing this right, then $f_{yyz}$ will equal $0$ and therefore $f_{yyzzx}$ will equal $0$.
Observe that $f(x, y, z) = x^2 \sin(4y) + z^3(6x - y) + y^4$ satisfies Clairaut's Theorem since it is the sum of products of smooth (infinitely continuously differentiable) functions. Consequently, the partial derivative $f_{yyzzx}$ can be computed via any permutation of the variables $y, y, z, z,$ and $x.$ Considering that $z$ only appears in one of the terms, it seems like the best option to begin. We find that $f_z = 3z^2(6x - y).$ Each of the variables appears in $f_z,$ so at this point, you could pick your favorite of these variables, and compute the partial derivative of $f_z$ with respect to it; however, as you can see, we must take the partial derivative with respect to $y$ twice, and the degree of $y$ in $f_z$ is 1. Ultimately, we conclude that $f_{yyzzx} = 0$ for all $x, y, z.$