To determine the equation (eqn.) of tangent (tgt.) line, say #t,# we
need, (1) slope of #t# and (2) a point (pt.) on #t.#
We already have a pt. #P(1,2) in t.#
To find the slope, let us recall that slope tgt. to a curve #gamma# is
given by #dy/dx" at "P, i.e., [dy/dx]_P.#
So, let us start by differentiating the eqn. of the curve
#gamma : x^2y^2+x^3+3=y^3.#
#:.d/dx[x^2y^2+x^3+3]=d/dx[y^3].#
#:. x^2d/dx(y^2)+y^2d/dx(x^2)+d/dx(x^3)+d/dx(3)=d/dx(y^3).#
#:. x^2*d/dy(y^2)*d/dx(y)+y^2*2x+3x^2+0=d/dy(y^3).dy/dx.#
#:. x^2*2y*dy/dx+y^2*2x+3x^2=3y^2*dy/dx.#
#:. (2x^2y-3y^2)dy/dx=-(2xy^2+3x^2).#
#:. dy/dx=-(x(2y^2+3x))/(y(2x^2-3y)).#
#:. [dy/dx]_{P(1,2)}=-{1(2*2^2+3*1)}/{2(2*1^2-3*2)}=(-11)/(-8).#
To Sum up, the line #t# has slope #11/8# and, #P(1,2) in t.#
By the Slope-Point Form of line, we have, the eqn. of
# t : y-2=11/8(x-1), or, 8y-16=11x-11, i.e.,#
# t : 11x-8y+5=0.#
Enjoy Maths.!
