#### Answer

$$
z_x= \frac{2x}{x^2+y^2} , \quad z_y= \frac{2y}{x^2+y^2} .
$$

#### Work Step by Step

Recall that $(\ln x)'=\dfrac{1}{x}$
Since $ z=\ln(x^2+y^2)$, then by using the chain rule, we have
$$
z_x= \frac{2x}{x^2+y^2} , \quad z_y= \frac{2y}{x^2+y^2} .
$$