I will use the notation by Ramanujan with $E_2, E_4,E_6$ being denoted by $P, Q, R$. For clarity let us write
\begin{align}
P(q) & =1-24\sum_{n=1}^{\infty}\frac {nq^n} {1-q^n}\tag{1a}\\
Q(q) &=1+240\sum_{n=1}^{\infty}\frac{n^3q^n}{1-q^n}\tag{1b}\\
R(q) &=1-504\sum_{n=1}^{\infty}\frac{n^5q^n}{1-q^n}\tag{1c}\end{align}
where $q=\exp(2\pi i\tau) $ and $\tau$ has positive imaginary part. Let us also list Ramanujan's identities dealing with derivatives of these functions
\begin{align}
q\frac{dP} {dq} &=\frac{P^2-Q}{12}\tag{2a}\\
q\frac{dQ}{dq}&=\frac{PQ-R}{3}\tag{2b}\\
q\frac{dR}{dq}&=\frac{PR-Q^2}{2}\tag{2c}
\end{align}
Further let us observe that $$\frac{dq} {d\tau} =2\pi iq$$ and hence $$\frac{d} {d\tau} =2\pi i q\frac{d} {dq} \tag{3}$$ We have by definition $$j(\tau) =\frac{12^3Q^3}{Q^3-R^2}\tag{4}$$ and one can prove using the equations $(2a),(2b),(2c)$ that $$q\frac{d} {dq} (Q^3-R^2)=P(Q^3-R^2)\tag{4}$$ Hence we have $$j'(\tau) =12^3\cdot 2\pi i q\frac{d} {dq} \frac{Q^3}{Q^3-R^2}=12^3(2\pi i) \cdot \dfrac{(Q^3-R^2)3Q^2q\dfrac{dQ}{dq}-Q^3q\dfrac{d}{dq}(Q^3-R^2)} {(Q^3-R^2)^2}=-12^3(2\pi i) \cdot\frac{Q^2R}{Q^3-R^2}$$ Differentiating again we get $$j''(\tau) =12^2(8\pi^2)\cdot\frac{PQ^2R-4QR^2-3Q^4}{Q^3-R^2}\tag{5}$$ One more round of differentiation gives us $$j'''(\tau) =12(64\pi^3 i)\cdot\frac{4R^3+\text{ terms containing }Q }{Q^3-R^2}\tag{6}$$ Next we use the link between $Q, R$ and elliptic integrals to evaluate them in closed form. Let $0<k<1$ be the elliptic modulus and $k'=\sqrt{1-k^2}$ be complementary modulus and let $K, K'$ be elliptic integrals defined by $$K=K(k) =\int_0^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}},K'=K(k')\tag{7}$$ Let $q'=\exp(-\pi K'/K) $ be the corresponding nome and then we have the standard formulas (proved here using the identities $(2a),(2b),(2c)$ mentioned above)
\begin{align}
Q(q'^2) &=\left(\frac{2K}{\pi}\right) ^4(1-k^2k'^2)\tag{8a}\\
R(q'^2) &=\left(\frac{2K}{\pi}\right)^6(1-2k^2)\left(1+\frac{k^2k'^2}{2}\right)\tag{8b}\\
Q(-q')&=\left(\frac{2K}{\pi}\right)^4(1-16k^2k'^2)\tag{8c}\\
R(-q')&=\left(\frac{2K}{\pi}\right)^6(1-2k^2)(1+32k^2k'^2)\tag{8d}
\end{align}
Further let us note that if $N$ is a positive rational number then there exists a unique positive value of modulus $k$ such that $K'/K=\sqrt{N} $ and this value $k$ is an algebraic number. Such $k$ are famously known as singular moduli. For small values of positive integer $N$ the values of $k=k_N$ as well as that of $K=K_N$ are well known. For our purposes we need to deal with $N=1,3$ and we have $$k_1=\frac{1}{\sqrt{2}},K_1=\frac {\Gamma ^2(1/4)}{4\sqrt{\pi}},k_3=\frac{\sqrt{3}-1}{2\sqrt{2}},K_3=\frac{3^{1/4}\Gamma ^{3}(1/3)}{2^{7/3}\pi} \tag{9}$$ (see details in this thread).
We can now evaluate $j''(i), j'''(\zeta_3)$ using the above given formulas. Let $\tau=i$ so that $q=e^{-2\pi}$ and then we can use the formulas $(8a),(8b),(9)$ with $q' =e^{-\pi}, k=k_1,K=K_1$ and note that $R=0$ and using $(5)$ we get desired value of $j''(i) $.
Next we use $\tau=\zeta_3$ so that $q=-e^{-\pi\sqrt{3}}$ and we use $(8c),(8d),(9)$ with $q' =e^{-\pi\sqrt{3}},k=k_3,K=K_3$ and note that $Q=0$ and using $(6)$ we get the desired value of $j''' (\zeta_3)$.
