The point (3,1) is on y = f^(-1) (x)

f(x) = x^3 +x+1 --->f'(x) = 3x+1 --->f'(1) = 4

-->the tangent line to the curve y = f^(−1) (x) at x = 3

is y-1 = (1/4)(x-3)

or y = (x+1)/4

y - 1 = m ( x - 3) where from [d(f^-1)/dx](b) = 1 / [( df/dx)](a) where ( a, b) is on the

graph of f(x)...here f(x) = x³ + x + 1 ===> df/dx = 3 x² + 1 and a = 1 when b = 3

thus m = 1 / 4...y - 1 = (1/4) [ x - 3 ] for your tangent line

Did you mean x^3 + x + 1 ?

dy/dx = 3x^2 + 1 =>

dx/dy = 1/(3x^2 + 1); at x = 3 this is 1/10.

The point (3,1) is on y = f^(-1) (x),

and the tangent line there is

y = (1/10)x + 7/10.