The slope of a tangent drawn to the graph of any function y = f(x) at a point where x = c is given by the value of f'(c).
If a line makes an angle of 45 degrees with the positive x-axis the slope of the line is tan 45 = 1.
The derivative of y = 3x^3 + 3x + 1 is y' = 9x^2 + 3
9x^2 + 3 = 1
=> 9x^2 = -2
=> x^2 = -2/9
This equation holds only if x is an imaginary number. At no point on the graph of the curve y = 3x^3 + 3x + 1 is the slope of the tangent 1. As verification consider the plot of y = 3x^3 + 3x + 1
There is no point where the tangent to y = 3x^3 + 3x + 1 makes an angle of 45 degrees with the positive x-axis.