Answer
${f_{xxyz}} = - \cos \left( {x + z} \right)$
Work Step by Step
We have $f\left( {x,y,z} \right) = y\sin \left( {x + z} \right)$.
Since $f$ is a continuous function, we can use Clairaut's Theorem in Section 15.3 to re-arrange the order of differentiation. So, ${f_{xxyz}} = {f_{yxxz}}$
${f_{xxyz}} = {f_{yxxz}} = \frac{\partial }{{\partial z}}\left( {\frac{\partial }{{\partial x}}\left( {\frac{\partial }{{\partial x}}\left( {\frac{{\partial f}}{{\partial y}}} \right)} \right)} \right)$
${f_{yxxz}} = \frac{\partial }{{\partial z}}\left( {\frac{\partial }{{\partial x}}\left( {\frac{\partial }{{\partial x}}\left( {\sin \left( {x + z} \right)} \right)} \right)} \right)$
$ = \frac{\partial }{{\partial z}}\left( {\frac{\partial }{{\partial x}}\left( {\cos \left( {x + z} \right)} \right)} \right)$
$ = \frac{\partial }{{\partial z}}\left( { - \sin \left( {x + z} \right)} \right)$
$ = - \cos \left( {x + z} \right)$
Thus, ${f_{xxyz}} = - \cos \left( {x + z} \right)$.
