The problem is:
$$ \begin{align} \operatorname{max} & \quad ax+by \\ \text{subject to} & \quad x+y=m. \end{align} $$
The Lagrangian is:
$$L(x,y) = ax+by−λ(x+y−m).$$
And so far I have:
$$ \begin{align} \frac{\partial L}{\partial x} & =a−λ=0, \\ \frac{\partial L}{\partial y} & =b−λ=0. \end{align} $$
How to solve it from here?
From the two equations you wrote at the end, you can see that $\lambda = a$ and $\lambda = b$. This means the problem can only be solved if $a=b$. In that case, the function you are minimizing is $ax+by=ax+ay=a(x+y)=a\cdot m$ constant (constant on the space where you are looking at).
This makes sense, since, if $a< b$ (a similar argument can be made for $a>b$), you can rewrite the function as $$ax+by = ax + by +ay - ay = a(x+y) + (b-a)y = am + (b-a)y.$$ Since you can now pick an arbitrary large value of $y$ and still have $x = m-y$ so that $x+y=m$, the function has no maximum.