Answer
$f_{xyy}=(f_{xy})_{y}$
$ f_{yxy}=(f_{yx})_{y}\qquad$... by the theorem, $f_{yx}$=$f_{xy}$, so
$(f_{xy})_{y}=(f_{xy})_{y}=f_{xyy}$
Thus, $f_{yxy}=f_{xyy}$
$ f_{yyx}=(f_{y})_{yx}\qquad$... by the theorem, $(f_{y})_{yx}$=$(f_{y})_{xy}$, so
$f_{yyx}=(f_{y})_{xy}=f_{yxy}$
Thus, $f_{yxy}=f_{yxy},$
By the transitive property of equality,
$f_{xyy}=f_{yxy}=f_{yyx}$
Work Step by Step
All steps are included in the answer.