#### Answer

$$y=2x-1$$

#### Work Step by Step

First we will find the derivative:
$$y'=\left(\sqrt{1+x^3}\right)'=\frac{1}{2\sqrt{1+x^3}}(1+x^3)'=\frac{3x^2}{2\sqrt{1+x^3}}$$
The tangent line of the curve at the point $(x_{0},y_{0})$ is given by formula:
$$y-y_{0}=y'(x_{0})(x-x_{0})$$
For the point $(x_{0},y_{0})=(2,3)$ we get:
$$y-3=\left.\frac{3x^2}{2\sqrt{1+x^3}}\right|_{x_{0}=2}(x-2)\Rightarrow y=\frac{3\cdot2^2}{2\sqrt{1+2^3}}(x-2)+3\Rightarrow y=\frac{12}{2\cdot3}(x-2)+3\Rightarrow y=2x-1$$