Building on Noam's suggestion, you could try using the inherent symmetry of the problem: Set $\sigma_1 = x^2 + y^2 + z^2$, $\sigma_2 = x^2y^2+y^2z^2+z^2x^2$, and $\sigma_3 = x^2y^2z^2$. Then your conditions become $\sigma_1 = c^2$ and $\sigma_2 = \sigma_3 + \tfrac14c^4$. Meanwhile, you have $$ dt^2 = dx^2 + dy^2 + dz^2 = \frac{d(x^2)^2}{4x^2}+\frac{d(y^2)^2}{4y^2}+\frac{d(z^2)^2}{4z^2}, $$ and this latter expression being symmetric in $x^2,y^2,z^2$, can be expressed as a differential expression in $\sigma_1, \sigma_2, \sigma_3$. I won't write out the details, but a short computation (using Maple) shows that, taking advantage of the relations $\sigma_1 = c^2$ and $\sigma_2 = \sigma_3 + \tfrac14c^4$, you come down to $$ dt^2 = \frac{\bigl(c^6(c^2{-}2)-4(36{-}52c^2{+}21c^4{-}2c^6)\sigma_3+16(c^2{-}1){\sigma_3}^2\bigr)}{16\sigma_3\bigl(c^6(2{-}c^2)-4(27{-}18c^2{+}2c^4)\sigma_3-16{\sigma_3}^2\bigr)}\bigl(d\sigma_3\bigr)^2 $$ Now, for example, you can see why $c^2=2$ is special. The integral that gives $t$ will simplify dramatically in this case; in fact, it becomes an elementary integral. For general values of $c$, though, this is a hyperelliptic integral, and you won't find any simple relation between $t$ and $\sigma_3$, so, a fortiori, none between $t$ and $x$, $y$, and $z$. There are various special values of $c$ for which the roots and poles of the rational expression cancel, such as $c=0$, $c = \pm\sqrt{2}$, and $c = \pm\frac32$, and, for these, you'd expect the integral to simplify considerably, but, otherwise, you don't expect any nice relation.

Building on Noam's suggestion, you could try using the inherent symmetry of the problem: Set $\sigma_1 = x^2 + y^2 + z^2$, $\sigma_2 = x^2y^2+y^2z^2+z^2x^2$, and $\sigma_3 = x^2y^2z^2$. Then your conditions become $\sigma_1 = c^2$ and $\sigma_2 = \sigma_3 + \tfrac14c^4$. Meanwhile, you have $$ dt^2 = dx^2 + dy^2 + dz^2 = \frac{d(x^2)^2}{4x^2}+\frac{d(y^2)^2}{4y^2}+\frac{d(z^2)^2}{4z^2}, $$ and this latter expression, being symmetric in $x^2,y^2,z^2$, can be expressed as a differential expression in $\sigma_1, \sigma_2, \sigma_3$. I won't write out the details, but a short computation (using Maple) shows that, taking advantage of the relations $\sigma_1 = c^2$ and $\sigma_2 = \sigma_3 + \tfrac14c^4$, this leads to the relation $$ dt^2 = \frac{\bigl(c^6(c^2{-}2)-4(36{-}52c^2{+}21c^4{-}2c^6)\sigma_3+16(c^2{-}1){\sigma_3}^2\bigr)}{16\sigma_3\bigl(c^6(2{-}c^2)-4(27{-}18c^2{+}2c^4)\sigma_3-16{\sigma_3}^2\bigr)}\bigl(d\sigma_3\bigr)^2. $$ Now, for example, you can see why $c^2=2$ is special. The integral that gives $t$ will simplify dramatically in this case; in fact, it becomes an elementary integral. For general values of $c$, though, this is a hyperelliptic integral, and you won't find any simple relation between $t$ and $\sigma_3$, so, a fortiori, none between $t$ and $x$, $y$, and $z$. There are various special values of $c$ for which the roots and poles of the rational expression cancel, such as $c=0$, $c = \pm\sqrt{2}$, and $c = \pm\frac32$, and, for these, you'd expect the integral to simplify considerably, but, otherwise, you don't expect any nice relation.

Added remark: By the way, you can get from this more directly to the relation between $t$ and $x$, $y$, and $z$, since, for example, letting $u$ represent any one of $x^2$, $y^2$, or $z^2$, one has the relation $u^3 - c^2 u^2 + (\sigma_3 + \tfrac14 c^4)u - \sigma_3 = 0$, which can clearly be solved for $\sigma_3$ as a rational function of $u$. Substituting this into the above relation gives a differential equation directly relating $t$ and, say, $u = x^2$. It's not a particularly nice relation, though. Ultimately, this gives a relation of the form $x = F(t,c)$ where $F$ is some function periodic of period $3\tau(c)$ in the first variable for some $\tau(c)>0$. Then one finds that $y = F(t + \tau(c),c)$ and $z = F(t-\tau(c),c)$. This is, of course, a very symmetric expression, though it's not explicit.