Implicit differentiation help

Use implicit differentiation to find the slope of the tangent line to the curve:

(y/(x+3y))=(x^2)+2

at the point ( 1, [3/(−8)] )

First find y' by using implicit differentiation (you'll need to use the quotient rule):
y / (x + 3y) = x² + 2
[y'(x + 3y) - y(1 + 3y')] / (x + 3y)² = 2x
y'(x + 3y) - y(1 + 3y') = 2x(x + 3y)²
xy' + 3yy' - y - 3yy' = 2x(x + 3y)²
xy' - y = 2x(x + 3y)²
xy' = y + 2x(x + 3y)²
y' = [y + 2x(x + 3y)²] / x

Now plug in the given point x = 1, y = -3/8:
y' = [-3/8 + 2(1)(1 + 3*-3/8)²] / 1
y' = -11/32

So, the slope of the tangent line at the given point is -11/32.

Hope that helps :)