Plug [tex]y=x^2[/tex] into the equation of the ellipsoid:
[tex]x^2+7(x^2)^2+7z^2=49[/tex]
Complete the square:
[tex]7x^4+x^2=7\left(x^4+\dfrac{x^2}7+\dfrac1{14^2}-\dfrac1{14^2}\right)=7\left(x^2+\dfrac1{14}\right)^2+\dfrac1{28}[/tex]
Then the intersection is such that
[tex]7\left(x^2+\dfrac1{14}\right)^2+7z^2=\dfrac{1371}{28}[/tex]
[tex]\left(x^2+\dfrac1{14}\right)^2+z^2=\dfrac{1371}{196}[/tex]
which resembles the equation of a circle, and suggests a parameterization is polar-like coordinates. Let
[tex]x(t)^2+\dfrac1{14}=\sqrt{\dfrac{1371}{196}}\cos t\implies x(t)=\pm\sqrt{\sqrt{\dfrac{1371}{196}}\cos t-\dfrac1{14}}[/tex]
[tex]y(t)=x(t)^2=\sqrt{\dfrac{1371}{196}}\cos t-\dfrac1{14}[/tex]
[tex]z=\sqrt{\dfrac{1371}{196}}\sin t[/tex]
(Attached is a plot of the two surfaces and the intersection; red for the positive root [tex]x(t)[/tex], blue for the negative)
