Answer
The equation of tangent line to the curve at $(2,3)$ is $$(l):y=\frac{1}{2}x+2$$
Work Step by Step
$$y=x+\frac{2}{x}$$$$y=x+2x^{-1}$$
1) Find the derivative of $y$ $$y'=\frac{d}{dx}(x)+2\frac{d}{dx}(x^{-1})$$
$$y'=1+2\times(-1)x^{-2}$$
$$y'=1-2x^{-2}$$
$$y'=1-\frac{2}{x^2}$$
2) Find the slope of the tangent line to the curve at $(2,3)$, or in fact, $y'(2)$ $$y'(2)=1-\frac{2}{2^2}$$$$y'(2)=1-\frac{1}{2}=\frac{1}{2}$$
3) The tangent line to the curve at $(2,3)$ would have the following equation: $$(l):y=y'(2)x+b$$$$y=\frac{1}{2}x+b$$
Since $(2,3)$ lies in the tangent line $(l)$, we can use that point to find $b$. In detail, $$\frac{1}{2}\times2+b=3$$$$1+b=3$$$$b=2$$
In conclusion, the equation of tangent line to the curve at $(2,3)$ is $$(l):y=\frac{1}{2}x+2$$
