Cauchy problem $$y z_x-xz_y=0$$ and $ x_0(s)=cos (s),y_0(s)=sin(s),z_0(s)=1,s>0$
I use Lagrange's method
$$\frac{dx}{y}=\frac{dy}{-x}=\frac{dz}{0}$$ From 1st and 2nd $x^2+y^2=c_1$ and from last relation $z=c_2$ So solution is of the type $z = f(x^2+y^2)$ By initial condition
$$1=f(cos^2s+sin^2s) \implies 1 = f(1) $$ so what i concluded from this about the solution to the problem ? Please help .
As you correctly found the general solution is $$z(x,y)=f(x^2+y^2)$$ and the condition is $$f(1)=1$$ Thus any function $f(X)$ which is equal to $1$ when the argument is equal to $1$ is convenient.
They are an infinity of functions such as $f(1)=1$ . Thus the specified condition is not sufficient to determine a unique solution. With the actual wording the problem has an infinity of solutions.
For examples :
With $f(X)=X^n$ then $z=(x^2+y^2)^n$ is solution.
With $f(X)=e^{X-1}$ then $z=e^{x^2+y^2-1}$ is solution.
With $f(X)=\sin(\frac{\pi}{2} X)$ then $z=\sin(\frac{\pi}{2} (x^2+y^2))$ is solution.
Etc.