we have the equation of curve C

Part a
Find out the gradient, where x=3
To find the gradient, take the derivative of the function with respect to x, then substitute the x-coordinate of the point of interest for the x values in the derivative
so
the first derivative is equal to

Evaluate the first derivative for x=3

the gradient is equal to 16
Part b
we know that the gradient is equal to 1 at point P
so
equate the first derivative to 1

Find out the possible y-coordinate of point P
For x=2
substitute in the given equation of C

the first possible coordinate of P is (2,-13)
For x=-2

the second possible coordinate of P is (-2,13)