Consider the IVP $$ y'=x\ln y, \quad y(1)=1 $$ The function $f(x,y)=x\ln y \,$ is continuous for $y>0$, so there exists a solution in some rectangle $$ R=\Big\{(x,y):|x-1|\leq a, |y-1|\leq b<1\Big\}$$
To show the uniqueness of this solution we take $$ f_y=\frac{\partial f(x,y)}{\partial y}=\frac{x}{y} $$ which is continuous in $\mathbb{R^2}\setminus \big\{\ (x,y): y=0 \big\}$.
Can we thus conclude that the IVP has a unique solution in some rectangle $$ R=\Big\{(x,y):|x-1|\leq a, |y-1|\leq b<1\Big\}$$
?