How do you find the slope of the tangent to the curve ##y^3x+y^2x^2=6## at ##(2,1)##?

Use and the

##3y^2dy/dxx+y^3+2ydy/dxx^2+y^2 2x=0 ##

Do some rewriting

##3xy^2dy/dx+2x^2ydy/dx+y^3+2xy^2=0##

Factor and move terms without a ##dy/dx ## factor to right side

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

now divide both sides by ##3xy^2+2x^2y ## and factor where you can

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

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

Now evaluate at the given point ## (2,1)##

##dy/dx=(-1(1+2(2)))/(2(3(1)+2(2)))=(-1(5))/(2(7))=(-5)/14 ##