Let $f\colon X \to Y$ be a Fréchet differentiable map between Hilbert spaces such that $f(0) = 0$ and $f$ is positively homogeneous. Then $f'(x)(z) = f(z)$ since $f(0+tz)-f(0) = tf(z)$ and so $f$ is a linear map.
Is this correct? This seems strange to me.
It is almost correct, but I would be slightly more careful/explicit with the argument, since you only assumed positive homogeneity; i.e for all $\xi \in X$ and for all $t>0$, $f(t\xi) = t f(\xi)$. If you assumed this for all $t\in \Bbb{R}$, then what you said is almost correct (you need to put $x=0$).
So, here's how I would go about it (it's pretty much the same idea). We'll show that for every $\xi \in X$, we have $f(\xi) = df_0(\xi)$, so that since $df_0$ is assumed to exist, and by definition is linear, we will have shown that $f$ is also linear. Now, for any $\xi \in X$, define the map $\gamma: \Bbb{R} \to X$ by $\gamma(t) = t \xi$. Now, $\gamma$ is differentiable at the origin, so the composite mapping $f \circ \gamma$ is also differentiable at the origin, so by the chain rule, \begin{align} (f\circ \gamma)'(0) &= df_{\gamma(0)} \big( \gamma'(0)\big) \\ &= df_0(\xi) \tag{$*$} \end{align}
By the way, take note that I'm using $df_0$ to mean the Frechet derivative (a linear map), whereas I'm using the $'$ notation, like $\gamma'(0)$ to mean the typical limit of the difference quotient \begin{align} \lim_{t \to 0} \dfrac{\gamma(t) - \gamma(0)}{t} \end{align} so that $\gamma'(0)$ is actually an element of $X$ rather than a linear map.
Having said this, what $(*)$ shows is that the (full double sided) limit $(f \circ \gamma)'(0)$ exists. Therefore, the typical one-sided limit $t \to 0^+$ also exists, and is equal to the double-sided limit. Hence, \begin{align} df_0(\xi) &= (f \circ \gamma)'(0) \\\\ &= \lim_{t \to 0^+} \dfrac{f(\gamma(t)) - f(\gamma(0))}{t} \\ &= \lim_{t \to 0^+} \dfrac{f(t \xi) - f(0)}{t} \\ &= \lim_{t \to 0^+} \dfrac{t f(\xi) - 0}{t} \tag{since $t>0$} \\ &= f(\xi) \end{align}
Since this is true for all $\xi \in X$, this completes the proof.
Extra Remarks:
A remark just for emphasis is that for this statement to be, it is absolutely essential that $f$ is differentiable at the origin, because otherwise this is false. Also, there is no need to assume the spaces $X,Y$ are Hilbert spaces, everything works if they're simply Banach spaces (the inner product structure was not relevant to this problem).
Also, take note that having proven this statement, we can actually conclude that $f$ is homogeneous (positive and negative), so that for all $t \in \Bbb{R}$, $f(t \xi) = t f(\xi)$. Therefore, positive homogeneity, $f(0) = 0$ and existence of $df_0$ together imply that $f$ is linear (and hence trivially homogeneous for all $t \in \Bbb{R}$).
