Compute [tex]\dfrac{\mathrm dy}{\mathrm dx}[/tex] via implicit differentiation:
[tex]x^2y+xy^2=3x\implies\dfrac{\mathrm d}{\mathrm dx}[x^2y+xy^2]=\dfrac{\mathrm d}{\mathrm dx}[3x][/tex]
[tex]\dfrac{\mathrm d}{\mathrm dx}[x^2y]+\dfrac{\mathrm d}{\mathrm dx}[xy^2]=3[/tex]
[tex]\left(\dfrac{\mathrm d}{\mathrm dx}[x^2]y+x^2\dfrac{\mathrm d}{\mathrm dx}[y]\right)+\left(\dfrac{\mathrm d}{\mathrm dx}[x]y^2+x\dfrac{\mathrm d}{\mathrm dx}[y^2]\right)=3[/tex]
[tex]\left(2xy+x^2\dfrac{\mathrm dy}{\mathrm dx}\right)+\left(y^2+2xy\,\dfrac{\mathrm dy}{\mathrm dx}\right)=3[/tex]
[tex](x^2+2xy)\dfrac{\mathrm dy}{\mathrm dx}=3-2xy-y^2[/tex]
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{3-2xy-y^2}{x^2+2xy}[/tex]
The value of [tex]\dfrac{\mathrm dy}{\mathrm dx}[/tex] at any point [tex](x,y)[/tex] on the given curve is the slope of the line tangent to it at this point. So at (-1, 1), we get
[tex]\dfrac{\mathrm dy}{\mathrm dx}=-4[/tex]
Then the equation of this tangent line is
[tex]y-1=-4(x-(-1))\implies y=-4x-3[/tex]
