I am using method described here and here.
With curve $\left(x^2+y^2+4 y\right)^2-16 \left(x^2+y^2\right)=0$ it works for me.
The curve has singular points $((0,0,1),(-i,1,0),(i,1,0))$ and I used additional ordinary point $(4, 0, 1)$.
Substituting these points into $a_1 x^2 + a_2 y^2 + a_3 z^2 + a_4 x y + a_5 x z + a_6 y z$ I eliminated some coefficients to get $a_1 x^2 + a_1 y^2 - 4 a_1 x z + a_6 y z$.
Then I computed resultants for $z=1$.
$$\operatorname{res}_x(a_1 x^2+a_1 y^2-4 a_1 x z+a_6 y z,(x^2+4 y+y^2)^2-16 (x^2+y^2))\\=y^3 (-128 a_1^2 a_6^2+32 a_1 a_6^3+1024 a_1^4 y-512 a_1^3 a_6 y+128 a_1^2 a_6^2 y-16 a_1 a_6^3 y+a_6^4 y)$$ $$\operatorname{res}_y(a1 x^2+a1 y^2-4 a1 x z+a6 y z,(x^2+4 y+y^2)^2-16 (x^2+y^2))\\=(-4+x) x^2 (-32 a1 a6^3+4 a6^4+1024 a1^4 x-512 a1^3 a6 x+128 a1^2 a6^2 x-16 a1 a6^3 x+a6^4 x)$$
We can see that there are factors that are linear in $y$ and $x$ which can be used to compute parametrization.
When I do the same with curve $y^4+8 x y^3-17 x^2 y^2-x y^2+8 x^3 y-2 x^2+5 x^3-2 x^4=0$ the resultants have never got any of the factors that would be linear in $x$ and $y$ no matter what combinations of points I used.
I used singular points $((0,0,1),(1,1,1)$ and different combinations of one to four ordinary points. Some of ordinary points: $$((1/2,0),(1/13,-(11/13)),(-(1/19),9/19),(1/37,7/37),(1/46,-(7/23)),(-(1/59),-(7/59)),(-(1/71),11/71),(2,0),(-(4/17),38/17),(16/151,52/151),(49/13,7/13),(49/109,7/109),(81/101,-(9/101)),(-(98/127),952/127),(121/2,-594),(121/109,-(11/109)),(121/193,-(11/193)),(169/34,26/17),(196/151,-(14/151)))$$
Am I doing any mistake or the method of adjoint curves does not work with all genus zero curves?