In the Proof for Existence of an Unique Solution (for Differential Equations), why $\frac{\partial f}{\partial y}$?

Theorem: Let $R$ be a rectangular region in the $xy$-plane defined by $[a,b]\times[c,d]$ that contains the point $(x_0,y_0)$ in its interior. If $f(x,y)$ and $\dfrac{\partial f}{\partial y}$ are continuous on $R$, then there exists some interval $I_0: (x_0-h,x_0+h), h>0$, which is contained in $[a,b]$, and a unique function $g(x)$, defined on $I_0$, which is a solution of $y'=f(x,y)$, given the initial condition $y(x_0)=y_0$.

I am taking an introductory differential equations course, and my professor introduced the above theorem today. My question is why do we have to check if $\dfrac{\partial f}{\partial y}$ is continuous instead of $\dfrac{\partial f}{\partial x}$? Can anyone explain why, keeping in mind I just started learning about differential equations about a week ago? Thanks in advance for any help.

• Probably because of an application of the inverse function theorem. – abnry Jan 8 '14 at 16:57
• That's just the Picard–Lindelöf theorem (it's something weaker, actually). It's hard to understand your question. If $\dfrac{\partial f}{\partial y}$ is continuous, it follows from the above theorem. If $\dfrac{\partial f}{\partial x}$ is continuous, it's probably not even true. Are you asking for a proof of the theorem, is that it? – Git Gud Jan 8 '14 at 17:01