On $x^3+y^3=z^3$, the Dixonian elliptic functions, and the Borwein cubic theta functions The Borwein cubic theta functions $a(q),b(q),c(q)$ and the doubly-periodic Dixonian elliptic functions $\mathrm{sm}(z), \mathrm{cm}(z)$ parameterize the Fermat cubic $x^3+y^3=1$, so
$$\frac{b^3(q)}{a^3(q)}+\frac{c^3(q)}{a^3(q)} =1\tag1$$
$$\mathrm{sm}^{3}(z)+\mathrm{cm}^{3}(z)=1\tag2$$
This post asks if there is any relation between the Borwein and Dixonian versions. It is known that the former is associated with $_2F_1\big(\tfrac13,\tfrac23;\color{blue}1;u\big)$ while the latter is with $_2F_1\big(\tfrac13,\tfrac23;\color{blue}{\tfrac43};v\big)$.
For argument $q=e^{2\pi i \sqrt{-n}}$, we have the nice relationship,
$$\large \frac{_2F_1\Big(\tfrac13,\tfrac23;1;\tfrac{b^3(q)}{a^3(q)} \Big)}{_2F_1\Big(\tfrac13,\tfrac23;1;\tfrac{c^3(q)}{a^3(q)} \Big)}=\sqrt{3n}$$

Q: By analogy, does the ratio,
  $$\frac{_2F_1\Big(\tfrac13,\tfrac23;\tfrac43;\mathrm{sm}^3(z) \Big)}{_2F_1\Big(\tfrac13,\tfrac23;\tfrac43;\mathrm{cm}^3(z) \Big)}=\,??\tag3$$
  evaluate to anything "meaningful" for appropriate choice of argument $z$?


P.S. To test this conjecture, for the circle $x^2+y^2=1$, we have the Dedekind eta quotients $\alpha(q), \beta(q)$ parameterization (given here) and the singly-periodic trigonometric functions,
$$\alpha^8(q)+\beta^8(q)=1\tag4$$
$$\sin^2 z+\cos^2 z=1\tag5$$
The analogous ratios,
$$\frac{_2F_1\Big(\tfrac12,\tfrac12;1; \beta^8(q) \Big)}{ _2F_1\Big(\tfrac12,\tfrac12;1; \alpha^8(q)\Big)}=\sqrt{4n}$$
and choice of $0<z<1$,
$$\frac{_2F_1\Big(\tfrac12,\tfrac12;\tfrac32; \cos^2 z \Big)}{ _2F_1\Big(\tfrac12,\tfrac12;\tfrac32; \sin^2 z \Big)}=\Big(\frac{\pi}{2z}-1\Big) \tan z$$
so $(3)$ might indeed evaluate to some closed-form, perhaps involving the fundamental constant of Dixonian functions $\large \pi_3=\frac{\sqrt3}{2\pi}\Gamma^3\big(\tfrac13\big)$.
 A: Recall that
$$\int_0^z\frac{\mathrm du}{\left(1-u^3\right)^{2/3}}=\operatorname{sm}^{(-1)}(z)=z\;{}_2 F_1\left({{\frac13,\frac23}\atop{\frac43}}\middle|z^3\right)$$
and
$$\int_z^1\frac{\mathrm du}{\left(1-u^3\right)^{2/3}}=\operatorname{cm}^{(-1)}(z)$$
along with
$$\frac{\pi_3}{3}=\int_0^1\frac{\mathrm du}{\left(1-u^3\right)^{2/3}}=\frac13 B\left(\frac13,\frac13\right)$$
so within a limited domain, your expression ought to simplify to
$$\frac{{}_2 F_1\left({{\frac13,\frac23}\atop{\frac43}}\middle|\operatorname{sm}^3(z)\right)}{{}_2 F_1\left({{\frac13,\frac23}\atop{\frac43}}\middle|\operatorname{cm}^3(z)\right)}=\frac{\frac{z}{\operatorname{sm}(z)}}{\frac{\frac{\pi_3}{3}-z}{\operatorname{cm}(z)}}=\frac{z}{\frac{\pi_3}{3}-z}\frac{\operatorname{cm}(z)}{\operatorname{sm}(z)}$$
Note the phrase "limited domain". This is expected, in the same way that $\arcsin\sin z=z$ and $\arccos\cos z=z$ are also only valid within a limited domain. In particular, the relation only works within the intersection of the domains of validity of $\operatorname{sm}^{(-1)}(\operatorname{sm}(z))$ and $\operatorname{cm}^{(-1)}(\operatorname{cm}(z))$ (both of which are, perhaps not surprisingly, hexagons), which is the rhombus with corners at $0,\frac{\pi_3}{3}\exp\left(\frac{\pi i}{3}\right),\frac{\pi_3}{3},\frac{\pi_3}{3}\exp\left(-\frac{\pi i}{3}\right)$.
