Introduction to Statistical Learning Eq. 6.12 and 6.13 Can someone please explain me how the optimization of 6.12 leads to 6.14 and the optimization of 6.13 leads to 6.15?

 A: For the first equation, it's the result of zero gradient;
$$
\begin{aligned}
S &= \sum_{j=1}^p (y_j-\beta_j)^2 +\lambda\sum_{j=1}^p\beta_j^2\\
\end{aligned}
$$
at extrema,
$$
\begin{aligned}
\frac{\partial S}{\partial \beta_j} &=0\\
-2(y_j -\beta_j) +2\lambda\beta_j &= 0\\
\beta_j &= \frac{y_j}{1+\lambda}.
\end{aligned}
$$
I think you should be able to derive the other expression using the same technique shown above and use the fact that
$$
\vert \beta_j \vert = \begin{cases} \beta_j \ \text{if} \ \beta_j > 0\\ -\beta_j \ \text{if} \ \beta_j < 0\end{cases}.
$$
A: The main technique has two steps: 1) take derivative w.r.t $\beta_j$ and 2) set it to $0$. I will outline the steps to help go from 6.12 to 6.14 and leave the other one.
Step 1):
Combining the summation and opening the quadratic expression of 6.12 gives
$$ \sum_{j=1}^p y_j^2 - 2y_j \times \beta_j + (1 + \lambda) \beta_j^2.$$
Now we take partial derivative of this entire summation w.r.t $\beta_j$ and get
$$ \frac{\partial}{\partial \beta_j} = -2 y_j + 2(1+\lambda) \beta_j. $$
Note the other $j'\neq j$ do not appear since they do no affect $\beta_j$.
Step 2):
We set the partial derivative to $0$ to find $\hat{\beta}_j^R$. Normally, we need to check the second order derivative (make sure it is positive) so that we know have found a minimizer. But we don't have to here because the quadratic form in $\beta_j$ is convex in $\beta_j$.
Barring some additional careful checking of the conditions on $y_j$, the Lasso regression (from 6.13 to 6.15) is very similar.
