Given the functions:

f(x) = x^3 + lnx and g(x) = 1/x + 3x^2

We need to verify if f'(x) = g(x).

First , we need to differentiate f(x).

We will use the differentiation identities to find the derivative

We know that:

f'(x) = (x^3 + lnx)'

==> f'(x) = (x^3)' + (lnx)'

= 3x^2 + 1/x

We notice that:

f'(x) = 3x^2 + 1/x = g(x).

**Then, f'(x) = g(x).**

The question gives f(x)=x^3+ln(x) and g(x)=1/x+3x^2.

Differentiating f(x), we use the relations that the derivative of x^n is n*x^(n-1) and the derivative of ln x is 1/x.

These five f'(x) = 3*x^2 + 1/x

This is the same as g(x).

**Therefore f'(x) = g(x).**

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