Calculus help, relatively simple but hard?

Well,

the curve's equation is:

x^2y + y = x + 3 ==> f(x,y) = x^2y + y - x - 3 = 0

we first verify that :

f(1,2) = 1^2 * 2 + 2 - 1 - 3 = 2 + 2 - 1 - 3 = 0 is on the curve : ok

then the differentiation gives :

2xy + x^2 y ' + y ' - 1 = 0

y ' * (x^2+1) = -2xy + 1

therefore :

dy/dx(1,2) = (-2*1*2 +1)/(1^2 +1)

the slope of the tangent is :

dy/dx(1,2) = -3/2

hope it' ll help !!

You're supposed to implicitly differentiate and solve for y'. I am not sure how you get 6 = 1? That means you never took the y' into account.

Differentiating implicitly, you should get:

x^2*y' + 2xy + y' = 1

Plugging the point in:

y' + 2(2) + y' = 1 => 2y' = -3 => y' = -3/2

This is a rational function:

y = (x+3)/(x^2+1), for all x.

You can use implicit differentiation or use the quotient rule and evaluate y'(x) at the point (1,2), since (1,2) is a point on the rational function y(x).