Question 554379
<pre>
6x² - 2xy + y³ = 9

Find the derivative implicitly term by term:

12x - 2(x{{{(dy)/(dx)}}} + y) + 3y²{{{(dy)/(dx)}}} = 0

Solve for {{{(dy)/(dx)}}}

12x - 2x{{{(dy)/(dx)}}} - 2y + 3y²{{{(dy)/(dx)}}} = 0

Isolate the terms in {{{(dy)/(dx)}}} 

-2x{{{(dy)/(dx)}}} + 3y²{{{(dy)/(dx)}}} = -12x + 2y

{{{(dy)/(dx)}}}(-2x + 3y²) = -12x + 2y

{{{(dy)/(dx)}}} = {{{(-12x+2y)/(-2x+3y^2)}}}

Write the positive terms first

{{{(dy)/(dx)}}} = {{{(2y-12x)/(3y^2-2x)}}}

{{{matrix(3,2,        "","",
           (dy)/(dx),"",
                 "", "(2,-3)") }}} = {{{(2(-3)-12(2))/(3(-3)^2-2(2))}}} = {{{(-6-24)/(3(9)-4)}}} = {{{(-30)/(27-4)}}} = {{{(-30)/23}}} = {{{-30/23}}}
    
Finding the tangent line is the algebra problem of finding the equation 
of the line through the point (2,-3) with a slope of {{{-30/23}}}

The equation of the tangent line at (2,-3) is

y = {{{-30/23}}}x - {{{9/23}}}

Finding the normal line is the algebra problem of finding the equation 
of the line through the point (2,-3) with a slope of {{{23/30}}}

The equation of the normal line at (2,-3) is

y = {{{23/30}}}x - {{{68/15}}}

Edwin</pre>