x^3+3xy+y^3=1 at the point (1,0)
i solved to get y =sqrt(x^3+3x-1)
then i think you take the derivative of that and set it equal to 0
im not sure how i procede
You need "implicit" differentiation.
$\displaystyle x^3+3xy+y^3=1$
Differentiate both sides, realising that the derivative of a constant is zero.
$\displaystyle 3x^2+3\frac{d}{dx}(xy)+\frac{d}{dx}y^3=0$
xy is a product and you need the chain rule for the "y-cubed" term.
Then, when you've got that, use x=1 and y=0.
The derivative at that point is the slope of the tangent.
Then the equation of the tangent is
$\displaystyle y-0=m(x-1)$
$\displaystyle y=\frac{dy}{dx}(x-1)$
is $\displaystyle 3y^2\frac{dy}{dx}$
Follow this example, you will find it very helpful.
Implicit Differentiation
your derivative is incorrect. it should be ...
$\displaystyle \frac{dy}{dx} = -\frac{x^2+y}{x+y^2}$
you need to review the method for taking implicit derivatives ...
Implicit Differentiation