Answer
\begin{aligned} \frac{d y}{d x}= -\frac{x+x^{2}+y^{2}}{y+x^{2}+y^{2}}\end{aligned}
Work Step by Step
Given $$\ln \sqrt{x^{2}+y^{2}}+x+y=4$$
$$ \Rightarrow \frac{1}{2}\ln (x^{2}+y^{2})+x+y-4=0$$
by letting $$F(x,y)=\frac{1}{2}\ln (x^{2}+y^{2})+x+y-4$$
So, we have
$$F_x(x,y)=\frac{\partial F(x,y)}{\partial x}= \frac{2x}{2(x^{2}+y^{2})}+1\\
\quad \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{x}{(x^{2}+y^{2})}+1=\frac{x+x^{2}+y^{2}}{(x^{2}+y^{2})} $$
$$F_y(x,y)=\frac{\partial F(x,y)}{\partial y}=\frac{2y}{2(x^{2}+y^{2})}+1\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{ y}{ (x^{2}+y^{2})}+1 =\frac{ y+x^{2}+y^{2}}{ (x^{2}+y^{2})}$$
Also, we get
\begin{aligned} \frac{d y}{d x}=&-\frac{F_{x}(x, y)}{F_{y}(x, y)} \\
&=-\frac{\frac{ x+x^{2}+y^{2}}{ (x^{2}+y^{2})}}{\frac{ y+x^{2}+y^{2}}{ (x^{2}+y^{2})}} \\ &=-\frac{x+x^{2}+y^{2}}{y+x^{2}+y^{2}}\end{aligned}