Answer
The second partial derivatives are as follows:
$f_{xx}=0;f_{yy}=\dfrac{2x}{y^3} ;\\ f_{xy}=f_{yx}=-\dfrac{1}{y^2}$
Work Step by Step
$$f(x,y)=y+\dfrac{x}{y}; \\f_{x}=\frac{1}{y}, f_{xx}=0, f_{yx}=-\dfrac{1}{y^2}; \\f_{y}=1-\frac{x}{y^2}, f_{yy}=\dfrac{2x}{y^3}, f_{xy}= f_{yx}$$
The second partial derivatives are as follows:
$f_{xx}=0;f_{yy}=\dfrac{2x}{y^3} ;\\ f_{xy}=f_{yx}=-\dfrac{1}{y^2}$
