Answer
As a result $z_x=\frac{1}{x+t^2}$, $z_t=\frac{2t}{x+t^2}$.
Work Step by Step
$z=\ln{(x+t^2)}$
In order to find $z_x$ treat $t$ as a constant and differentiate with respect to $x$.
$z_x=\frac{1}{x+t^2}$
In order to find $z_y$ treat $x$ as a constant and differentiate with respect to $t$.
$z_t=\frac{2t}{x+t^2}$