I've attached a plot of the intersection (highlighted in red) between the parabolic cylinder (orange) and the hyperbolic paraboloid (blue).
The arc length can be computed with a line integral, but first we'll need a parameterization for \(C\). This is easy enough to do. First fix any one variable. For convenience, choose \(x\).
Now, \(x^2=2y\implies y=\dfrac{x^2}2\), and \(3z=xy\implies z=\dfrac{x^3}6\). The intersection is thus parameterized by the vector-valued function
\(\mathbf r(x)=\left\langle x,\dfrac{x^2}2,\dfrac{x^3}6\right\rangle\)
where \(0\le x\le 4\). The arc length is computed with the integral
\(\displaystyle\int_C\mathrm dS=\int_0^4\|\mathbf r'(x)\|\,\mathrm dx=\int_0^4\sqrt{x^2+\dfrac{x^4}4+\dfrac{x^6}{36}}\,\mathrm dx\)
Some rewriting:
\(\sqrt{x^2+\dfrac{x^4}4+\dfrac{x^6}{36}}=\sqrt{\dfrac{x^2}{36}}\sqrt{x^4+9x^2+36}=\dfrac x6\sqrt{x^4+9x^2+36}\)
Complete the square to get
\(x^4+9x^2+36=\left(x^2+\dfrac92\right)^2+\dfrac{63}4\)
So in the integral, you can substitute \(y=x^2+\dfrac92\) to get
\(\displaystyle\frac16\int_0^4x\sqrt{\left(x^2+\frac92\right)^2+\frac{63}4}\,\mathrm dx=\frac1{12}\int_{9/2}^{41/2}\sqrt{y^2+\frac{63}4}\,\mathrm dy\)
Next substitute \(y=\dfrac{\sqrt{63}}2\tan z\), so that the integral becomes
\(\displaystyle\frac1{12}\int_{9/2}^{41/2}\sqrt{y^2+\frac{63}4}\,\mathrm dy=\frac{21}{16}\int_{\arctan(3/\sqrt7)}^{\arctan(41/(3\sqrt7))}\sec^3z\,\mathrm dz\)
This is a fairly standard integral (it even has its own Wiki page, if you're not familiar with the derivation):
\(\displaystyle\int\sec^3z\,\mathrm dz=\frac12\sec z\tan z+\frac12\ln|\sec x+\tan x|+C\)
So the arc length is
\(\displaystyle\frac{21}{32}\left(\sec z\tan z+\ln|\sec x+\tan x|\right)\bigg|_{z=\arctan(3/\sqrt7)}^{z=\arctan(41/(3\sqrt7))}=\frac{21}{32}\ln\left(\frac{41+4\sqrt{109}}{21}\right)+\frac{41\sqrt{109}}{24}-\frac98\)
