Chapter 14 - Section 14.3 - Partial Derivatives - 14.3 Exercise - Page 927: 97

No

Here, we have $f_{x}(x, y)=x+4y ; f_{xy}(x, y)=4$ and $ f_{y}(x, y)=3x-y \implies f_{yx}(x, y)=3$ Here, the second order derivatives are not equal and both $f_{xy}$ and $f_{yx}$ are continuous. By Clairaut's Theorem it should be that $f_{xy}(x, y)=f_{yx}(x, y).$ This implies that such a function $f$ does not exist.