Answer
${f_{xyxzy}} = 0$
Work Step by Step
Using Clairaut's Theorem, we arrange the order of the partial differentiations such that:
${f_{xyxzy}} = {f_{xxyyz}}$
So,
${f_x} = yz\cos \left( {xz} \right)\sin \left( {x + z} \right) + y\sin \left( {xz} \right)\cos \left( {x + z} \right) + \tan y + \tan \left( {\frac{{z + {z^{ - 1}}}}{{y - {y^{ - 1}}}}} \right)$
${f_{xx}} = - y{z^2}\sin \left( {xz} \right)\sin \left( {x + z} \right) + yz\cos \left( {xz} \right)\cos \left( {x + z} \right)$
$ + yz\cos \left( {xz} \right)\cos \left( {x + z} \right) - y\sin \left( {xz} \right)\sin \left( {x + z} \right)$
${f_{xxy}} = - {z^2}\sin \left( {xz} \right)\sin \left( {x + z} \right) + z\cos \left( {xz} \right)\cos \left( {x + z} \right)$
$ + z\cos \left( {xz} \right)\cos \left( {x + z} \right) - \sin \left( {xz} \right)\sin \left( {x + z} \right)$
${f_{xxyy}} = 0$
${f_{xxyyz}} = 0$
Therefore, ${f_{xyxzy}} = {f_{xxyyz}} = 0$
