Closed form for $f(z)^2 + f ' (z)^2 + f ' ' (z) ^2 = 1 $? Can we give a closed form for $f(z)$ in
$$ f(z)^2 + f ' (z)^2 + f ' ' (z) ^2 = 1 $$
Apart from $f(x)= 1$ or $f(x)= -1$.
Where " closed form " means in terms of standard functions , integrals and the inv operator ( e.g. $\text{Inv}( z \exp(z) ) = \text{LambertW}_0(x)$ ).
Other insights are also appreciated.
( example uniqueness , singularities , periodicity ... )
 A: $$  f(z) = 1  $$
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Th other answer has some complex sinusoidals that work. I was trying to make a series solution with real coefficients beginning with $Ax$ so it goes through the origin, with $A^2 + B^2 = 1$ real constants it begins
$$ Ax + \frac{B}{2} x^2   - \frac{A}{6} x^3    - \frac{1}{24B} x^4 \cdots   $$
In contrast, if you just take $y^2 + y'^2 = 1,$ then $y = \sin x$ works nicely.
A: *

*A remark: the equation does not contain independent variable and therefore admit a reduction of order. Indeed, let $\frac{df}{dz}=s$, $f=t$ and notice that $\frac{d^2f}{dz^2}=\frac{ds}{dt}\cdot\frac{dt}{dz}=s\frac{ds}{dt}$. The equation thus becomes
$$s^2 \left(\frac{ds}{dt}\right)^2+s^2+t^2=1.$$

*There are some "easy-to-guess" one-parameter families of solutions. For example, set
$$f(z)=A\sin(az+\phi),$$
so that 
$$f'(z)=Aa\cos(az+\phi),\qquad f''(z)=-Aa^2\sin(az+\phi).$$
This will obviously be a solution provided that
$$A^2 a^2=1,\qquad a^4+1=a^2. $$
Parameter $\phi$ can be arbitrary and plays the role of integration constant.
