I am trying to understand this equation given here to calculate Mean squared loss.
$$ \frac{\text{d}MSE}{\text{d}a} = \frac{\text{d}MSE}{\text{d}y} . \frac{\text{d}y}{\text{d}a} = \frac{1}{N}\sum_{i=1}^N 2(y - a -bx) . (-1) $$
Doing partial diff following the product chain rule, Here are the steps I took $$ \tag{1} \frac{\text{d}MSE}{\text{d}a} = \frac{\text{d}MSE}{\text{d}y} . \frac{\text{d}y}{\text{d}a}$$
$$ \tag{2} \frac{\text{d}MSE}{\text{d}y} = \frac{1}{N}\sum_{i=1}^N 2(y - a - bx) $$
The third partial diff should be (I am not sure about it)
$$ \tag{3} \frac{\text{d}y}{\text{d}a} = \frac{1}{N}\sum_{i=1}^N 2(-1) $$
Any suggestion where did I go wrong
The link is rather incompletely written, and even more incompletely reproduced in the question. There is some Einstein sum convention used but not mentioned, the assignment of inner derivatives is not clear,...
You get $$ MSE = \sum_{i=1}^N(y_i-\hat y_i)^2\implies \frac{∂ MSE}{∂\hat y_i}=2(y_i-\hat y_i)(-1)=2(\hat y_i-y_i) $$ as you have a square of a difference, and the "active" variable is the second term. Then $$ \hat y_i=a+bx_i\implies \frac{∂\hat y_i}{∂a}=1~~\land ~~\frac{∂\hat y_i}{∂b}=x_i $$ Then in combination, the intermediate variable in the cited chain rule is the full vector $\bf \hat y$ of all the $\hat y_i$, $$ \frac{∂ MSE}{∂a}=\frac{∂ MSE}{∂\bf\hat y}\frac{∂\bf\hat y}{∂a} =\sum_{i=1}^N\frac{∂ MSE}{∂\hat y_i}\frac{∂\hat y_i}{∂a} =\sum_{i=1}^N 2(a+bx_i-y_i)\cdot 1 =-2\sum_{i=1}^N (y_i-a-bx_i) , $$ $$ \frac{∂ MSE}{∂b}=\frac{∂ MSE}{∂\bf\hat y}\frac{∂\bf\hat y}{∂b} =\sum_{i=1}^N\frac{∂ MSE}{∂\hat y_i}\frac{∂\hat y_i}{∂b} =\sum_{i=1}^N 2(a+bx_i-y_i)\cdot x_i =-2\sum_{i=1}^N (y_i-a-bx_i) ⋅x_i $$ etc.