y^3+y^2-5y-x^2=-4

find the slope of the tangent line to the curve x^2+y^2=16 at x=3/

find the points on the graph y=x^2+1 at which the slope is 0.

Find the equation of the tangent line to the curve y=2x/1-x^2 at x=0.

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For solving any of the above problem we have to find derivative of the function dy/dx

Lets take

find the points on the graph y=x^2+1 at which the slope is 0.

dy/dx = 2x

We are given that slope is 0

dy/dx =0

or x =0

Putting this in the original function we have

 y=x^2+1

y=1

So point is (0 ,1 ) at which slope is 0

You can try others too similarly , just find derivative and put slope as in given question