Find the equation of the tangent to the curve x^3+yx^2=1 at the point (1,0).

We can find the gradient of a tangent to a curve at a point by finding dy/dx at x=1.Firstly we can rearrange the equation of the curve for y.(1) yx^2=1-x^3(2)y=x^{-2}-xThen we can differentiate the equation w.r.t x to find dy/dxdy/dx=-2x^{-3}-1At x=1 dy/dx=-2-1=-3To find the equation of a tangent at a point we can use y-y_{1}=m(x-x_{1})Inputting the values into this equation gives(1) y-0=-3(x-1)(2) y=-3x+3Which is the equation of the tangent to the curve at the point (1,0) so we're done.