[tex]x^y y^x=M[/tex]
[tex]\ln(x^y y^x)=\ln M[/tex]
[tex]y\ln x+x\ln y=\ln M[/tex]
Differentiating both sides with respect to [tex]x[/tex], and assuming [tex]y=y(x)[/tex], we have
[tex]\dfrac{\mathrm d}{\mathrm dx}[y\ln x+x\ln y]=0[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}\ln x+\dfrac yx+\ln y+\dfrac xy\dfrac{\mathrm dy}{\mathrm dx}=0[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}\left(\ln x+\dfrac xy\right)=-\dfrac yx-\ln y[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{\frac yx+\ln y}{\frac xy+\ln x}[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=-\dfrac{y^2+xy\ln y}{x^2+xy\ln x}[/tex]
In case you wanted the derivative with respect to [tex]y[/tex] under the assumption that [tex]x=x(y)[/tex], the result would be the same with the exception that the left hand side getting written as [tex]\dfrac{\mathrm dx}{\mathrm dy}[/tex]. This is due to the symmetry of the original left hand side of the equation:
[tex]f(y,x)=y^xx^y=x^yy^x=f(x,y)[/tex]
