Looks like the equation is
[tex]g(x) + x \cos(g(x)) = x^3[/tex]
Differentiate both sides with respect to x, using the chain rule on the left side:
[tex]g'(x) + \cos(g(x)) - x \sin(g(x)) g'(x) = 3x^2[/tex]
Solve for g'(x) :
[tex]g'(x) (1 - x \sin(g(x))) = 3x^2 - \cos(g(x)) \\\\ g'(x) = \dfrac{3x^2 - \cos(g(x))}{1 - x \sin(g(x))}[/tex]
When x = 0, we have
[tex]g'(0) = \dfrac{0-\cos(g(0))}{1-0\sin(g(0))} = \boxed{-\cos(g(0))}[/tex]
(This is all you can say without providing the value of g (0).)
