In special relativity we've the invariant $$ d s^2=-d t^2 +d x^2 + d y^2+d z^2 $$
For a clock moving along the worldline in question the above equation reduces to $\begin{aligned} d s^2=&-d t^2\end{aligned}$ , hence we can say that the time measured by the clock moving along the world line reads time $dt$ such that $\begin{aligned} d s^2=&-d t^2\end{aligned}$ which is called proper time.
Then Hartle gravity pg 126 while motivating curvature of space gives an example of geometry such that
$$d s^2=-\left(1+\frac{2 \Phi\left(x\right)}{c^2}\right)(c d t)^2+\left(1-\frac{2 \Phi\left(x\right)}{c^2}\right)\left(d x^2+d y^2+d z^2\right)$$
In this space I want an expression for the proper time.
Suppose in one frame I measure two events $A$ and $B$. I calculate the interval $ds$ using $d s^2=-\left(1+\frac{2 \Phi\left(x\right)}{c^2}\right)(c d t)^2+\left(1-\frac{2 \Phi\left(x\right)}{c^2}\right)\left(d x^2+d y^2+d z^2\right)$
Any other frame will measure the same interval.
Now suppose we take a clock moving along the worldline of our events. In the clocks frame the spatial differentials $dx, dy, dz$ for the two events $A$ and $B$ are zero. Suppose in the moving clocks frame we measure time $dt$.
This time $dt$ is also the proper time $d\tau$
To find proper time we put $dx, dy, dz=0$ in $d s^2=-\left(1+\frac{2 \Phi\left(x\right)}{c^2}\right)(c d t)^2+\left(1-\frac{2 \Phi\left(x\right)}{c^2}\right)\left(d x^2+d y^2+d z^2\right)$ which gives me the proper time as $$d{ \tau^2}=\frac{-d s^2}{c^2\left(1+\frac{2 \phi}{c^2}\right)}$$
but the author says that the proper time is $$d \tau^2=-d s^2 / c^2$$
Where am I wrong?