Answer
No.
Work Step by Step
$\left[\begin{array}{lll}
f_{x}(x, y)=x+4y & , & f_{y}(x, y)=3x-y\\
f_{xy}(x, y)=4 & & f_{yx}(x, y)=3
\end{array}\right]$
We have $f_{xy}$ and $f_{yx}$ both continuous, so, by Clairaut's Theorem it should be that
$f_{xy}(x, y)=f_{yx}(x, y).$
Since it is not so, such a function f does not exist.
