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Your belief is not correct, as Robert Israel pointed in his comment. Same for differential equations: $yy'-(3/2)z^2=0$ has a solution $y(z)=z^{3/2}$ which is not analytic at $0$. The correct theorem is: If $P(z_0,y_0,y_1,\ldots,y_{n})=0$, where $P$ is a polynomial, (or an analytic function in a neighborhood of $(z_0,\ldots,y_n)$), and $\partial P/\partial y_{n}\neq 0$ at the same point, then there exists a unique analytic function $y(z)$ in a neighbourhood of $z_0$, which satisfies $P(z,y,y',\ldots,y^{(n)})=0$ in this neighbourhood, and $y(z_0)=y_0,\; y'(z_0)=y_0,\ldots,y^{(n)}(z_0)=y_{n}$. One does not need to assume a priori that $y(z)$ is analytic. If it is $n$ times differentiable in a neighbourhood of $z_0$, it is automatically analytic. For the proof, see, for example H. Cartan, Elementary theory of analytic functions of one and several complex variables, or Coddington and Levinson, Ordinary differential equations.