Equation with partial derivative

Let $u=-(x+y)/2$ and $v=(3x+y)/2$. Then $x=u+v$ and $y=-3u-v$. Let $$F(u,v)=f(x(u,v),y(u,v))=f(u+v,-3u-v) $$ Then we have $$f(x,y)=F(u(x,y),v(x,y))=F(-(x+y)/2,(3x+y)/2) $$ Now we have $$\partial _u F(u,v)=(\partial x/\partial u)\partial_xf +(\partial y/\partial u)\partial_yf=\partial_xf -3\partial_yf =0$$ So $F$ does not depend on $u$ and we have $F(u,v)=g(v)$ for some function $g$. Therefore $$ f(x,y)=F(u(x,y),v(x,y))=g(v(x,y))=g((3x+y)/2)$$for an arbitrary continuously differentiable function of one variable $g$. This is the general solution and can be easily checked that it satisfies the equation. Since $g(t)$ is a function of one variable for $t=(3x+y)/2$ we have $$\partial _x f =g'(t)\partial_x t=3g'(t)/2$$ Similarly $\partial_y f=g'(t)/2$. Hence $$\partial_x f-3\partial_yf=3g'(t)/2-3g'(t)/2=0$$