I don't know why these two method yield different solutions for marginal product of labour Given a homogeneous of degree one function
$$Y=F(K,N) \\
y=\frac{Y}{N}
=F(\frac{K}{N},1):=f(k)$$
I am looking for the partial derivative for $N$
Method 1
$$F_N=\frac{\partial F(K,N)}{\partial N}=\frac{\partial Nf(k)}{\partial N}\\
F_N=f(k)+N\frac{\partial f(k)}{\partial N}\\
F_N=f(k)+N\frac{\partial f(k)}{\partial k}\frac{\partial k}{\partial N}, \quad k=KN^{-1}\\
F_N=f(k)+Nf'(k)(-1)KN^{-2}\\
F_N=f(k)+f'(k)(-1)KN^{-1}=f(k)-f'(k)k
 $$
Method 2
$$F_N=\frac{\partial F(K,N)}{\partial K}\frac{\partial K}{\partial N}\\
F_N=\frac{N\partial f(k)}{N\partial k}\frac{\partial K}{\partial N}\\
F_N=f'(k)\frac{\partial Nk}{\partial N}=f'(k)k
$$
I have no idea where is the method 2 gone wrong.  Any help would be appreciated!
Re-edited Solved
$$\frac{dF(K,N)}{dN}=\frac{\partial F(K,N)}{\partial K}\frac{dK}{dN}+\frac{\partial F(K,N)}{\partial N}\frac{dN}{dN}\\
\frac{dNf(k)}{dN}=f'(k)k+\frac{\partial F(K,N)}{\partial N}\\
f(k)-f'(k)k=\frac{\partial F(K,N)}{\partial N}
$$
I am not sure why you expect method two to work, as generally
$$
\frac{\partial F(K,N)}{\partial N} \neq \frac{\partial F(K,N)}{\partial K}\frac{\partial K}{\partial N}.
$$
These are partial derivatives. Unless information to the contrary exists
$$\frac{\partial K}{\partial N} = 0.$$
You can easily see that these formulas are not connected by calculating the area of a rectangle which sides have lengths of $a,b$. Thus $A(a,b) = a\cdot b$ and
$$
\frac{\partial A(a,b)}{\partial b} = a \neq b \cdot 0 = \frac{\partial A(a,b)}{\partial a}\frac{\partial a}{\partial b}.
$$
In case you insist on homogeneity of degree one you can take the square root of the area.