As we are seeking slope at #(1,4)#, this point lies on the curve given by #x^3y^2-(x+y)/(x-y)^2=C#.

Putting values from #(1,4)#, we get #1^3*4^2-(1+4)/(1-4)^2=C#

or #C=16-5/9=139/9# and function is #x^3y^2-(x+y)/(x-y)^2=139/9#

Now differentiating the imlicit function #x^3y^2-(x+y)/(x-y)^2=C#

#3x^2y^2+2x^3y(dy)/(dx)-(1+(dy)/(dx))/(x-y)^2-(2(x+y))/(x-y)^3(1-(dy)/(dx))=0#

or #3x^2y^2(x-y)^3+2x^3y(x-y)^3(dy)/(dx)-(x-y)(1+(dy)/(dx))-2(x+y)(1-(dy)/(dx))=0#

or #[2x^3y(x-y)^3-(x-y)+2(x+y)] (dy)/(dx)=[-3x^2y^2(x-y)^3+x-y+2(x+y)]#

or #(dy)/(dx)=[-3x^2y^2(x-y)^3+x-y+2(x+y)]/[2x^3y(x-y)^3-(x-y)+2(x+y)]#

= #[-3x^2y^2(x-y)^3+3x+y]/[2x^3y(x-y)^3+x+3y]#

and at #(1,4)#, slope is #[-3*1^2*4^2(1-4)^3+3+4]/[2*1^3*4(1-4)^3+1+3*4]#

= #(48*27+7)/(-8*27+13)#

= #-1303/203#

The tangent at #(1,4)# looks liike:

graph{(x^3y^2-(x+y)/(x-y)^2-139/9)(203y-812+1303x-1303)=0 [-10.87, 9.13, -4.08, 5.92]}