Let's first prove that ideal $I:=(x^2-y, x^3-z)$ is prime. Suppose $f\cdot g \in I$. Using obvious isomorphisms $k[x,y,z] \cong (k[x,y])[z]$, $k[x,y] \cong (k[x])[y]$ and division algorithm, we have
$$ f(x,y,z)=(x^3-z)f_1(x,y,z) + (x^2-y)f_2(x,y) + f_3(x), $$
$$ g(x,y,z)=(x^3-z)g_1(x,y,z) + (x^2-y)g_2(x,y) + g_3(x). $$
Now we have $f_3(x) \cdot g_3(x) \in I$, therefore
$$f_3(x) \cdot g_3(x)=(x^3-z)h_1(x,y,z) + (x^2-y)h_2(x,y,z).$$
Insert $(x,y,z)=(t,t^2,t^3)$ and get $f_3(t) \cdot g_3(t) =0$ for all $t \in k$. If $k$ is algebraically closed (therefore infinite), we have $f_3 \cdot g_3 = 0$, so $f_3 = 0$ or $g_3= 0$. Then $f \in I$ of $g \in I$, so $I$ is prime (and therefore radical).
We have $I(Y)=I(V(I)) = \operatorname{Rad}(I) = I$, which is prime. So $Y$ is irreducible.
As I explained it here one way is to first find the Hilbert function $h_C(d)$ and see what polynomial it is when $d \gg 0.$
Let $H=\{w=0\} \subset \mathbb{P}^3$ and consider $X=C \cap H=V(w,xz-y^2,z^2,yz).$ Then the homogeneous coordinate ring of $X$ is (isomorphic to) $S(X)=k[x,y,z]/(xz-y^2,z^2,yz).$
Exercise: Show that its degree $d$ part, $S(X)^{(d)},$ is generated by $x^d,x^{d-1}y, x^{d-1}z$ as a vector space over $k,$ and conclude that $h_X(d)=\dim_k S(X)^{(d)}=3,$ so is $p_X(d)=3.$
Now the question is, how to relate $C$ and $X?$
The answer is the following SES
$$0\longrightarrow k[x,y,z,w]/I \stackrel{.w}{\longrightarrow} k[x,y,z,w]/I \longrightarrow k[x,y,z,w]/(I+(w)) \longrightarrow 0$$
where $I=(y^2-xz,z^2-yw,xw-yz).$ (It would be easy to run the cohomological approach if you like, using this SES.)
Take the $d$ graded pieces of the SES to get $h_C(d)=h_C(d-1)+h_X(d)$ and note that $h_C(0)=1,$ since $S(C)^{(0)}=k.$ Finally, prove that $h_C(d)=3d+1$ which is a polynomial, thus $p_C(d)=3d+1.$
In general, the degree of the Hilbert polynomial of a projective subscheme $X$ of $\mathbb{P}^n$ coincides with the dimension of $X$, so there should be no surprise that why in this case is a degree one polynomial.
Moreover, the degree of $X$ is defined to be $(\dim X)! \times \text{leading coefficient of} \; p_X(d),$ which in our case is $3.$
Best Answer
An alternative description is that (x,y,z) belongs to the tangent surface if and only if the polynomial (in the variable $s$) $1+3sx+3s^2y+s^3z$ has a double root. The corresponding M2 commands to get the equation of the tangent surface are
R=QQ[s,x,y,z]
factor (discriminant(1+3*s*x+3*s^2*y+s^3*z,t))
The reason is that the twisted cubic corresponds to polynomials $p(t)=(1+st)^3$ and the tangent line at $p(t)$ is parametrized by cubic polynomials $p(t)+up'(t)=(1+st)^2(1+(t+3u)s)$.