Answer
No
Work Step by Step
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.