Answer
We use Clairaut's Theorem and show that ${f_{xyx}}$ also exists.
Work Step by Step
Since ${f_{xy}}$ and ${f_{yx}}$ are continuous, by Clairaut's Theorem:
(1) ${\ \ \ }$ ${f_{xy}} = {f_{yx}}$
Since ${f_{yxx}}$ exists, it implies that the derivative of ${f_{yx}}$ with respect to $x$ exists. Thus, the derivatives of equation (1) with respect to $x$ becomes ${f_{xyx}} = {f_{yxx}}$.
Hence, ${f_{xyx}}$ also exists.
