From the comments above:
QUOTE: Sorry, this is so confusing, I'm probably not explaining myself clearly. I'm not in the mathematics field. So, I've seen a paper that shows the partial derivative of $F(k, λ)$ with respect to $λ$, and their conclusion was that as $λ$ increases and $k$ is held constant, $F$ decreases. I want to do the same but for the $k$. END OF QUOTE
Since $k$ takes on only the values $\{0,1,2,3,\ldots\}$ and does not vary continuously, the thing to do here is to use finite differences rather than derivatives. I.e. instead of asking how fast the function is changing as $k$ varies, ask how much the function changes as $k$ increases by $1$.
So you have
$$
\begin{align}
& {} \qquad \frac{\lambda^{k+1}e^{-\lambda}}{(k+1)!} - \frac{\lambda^k e^{-\lambda}}{k!} = e^{-\lambda} \lambda^k\left(\frac{\lambda}{(k+1)!} - \frac{1}{k!}\right) \\ \\
& = e^{-\lambda} \lambda^k\left(\frac{\lambda}{(k+1)!} - \frac{k+1}{(k+1)!}\right) = \frac{e^{-\lambda} \lambda^k}{(k+1)!}(\lambda-k-1).
\end{align}
$$
This is positive when $k<\lambda -1$ and negative when $k>\lambda-1$, and if $\lambda$ happens to be an integer, it is $0$ when $k=\lambda-1$.
Therefore the function increases on $\{0,1,2,3,\ldots,\lfloor\lambda-1\rfloor\}$ and decreases on $\{\lfloor\lambda-1\rfloor+1,\lfloor\lambda-1\rfloor+2,\lfloor\lambda-1\rfloor+3,\ldots\}$.
You might still need to look at the question of which of two values is bigger. In that one isolated case of an exact ineger value, the maximum is attained at each of two consecutive integers.
The quantity $\frac{\partial f(x,y)}{\partial x}$ is defined as the derivative of $f$ with respect to its first argument. Even though you will eventually assign $y = \phi(x)$, the quantity $\frac{\partial f(x,y)}{\partial x}$ still refers to the (perhaps hypothetical) assessment of "how much does $f$ vary when $x$ is varied and $y$ (its second argument) is held constant." Meanwhile, the total derivative $\frac{df(x,y)}{dx}$ assesses all ways that $f$ varies with $x$, including any "indirect" contribution through how $y$ may depend on $x$ (i.e. $\phi$). Generally speaking,
$$
\frac{df(x,y)}{dt} = \frac{\partial f}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial t}
$$
Letting $x=t$ we have,
\begin{align}
\frac{df(x,y)}{dx} &= \frac{\partial f}{\partial x} \frac{\partial x}{\partial x} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial x}\\
&= \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \frac{\partial y}{\partial x}
\end{align}
Letting $y = \phi(x)$ we have,
$$
\frac{df(x,y)}{dx} = \frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} \phi'(x)
$$
Notice that,
$$
\phi'(x)=0\ \implies\ \frac{df}{dx}=\frac{\partial f}{\partial x}
$$
The partial derivative under question is "how $f$ varies while we vary $x$ and hold its other arguments constant", $\frac{\partial f}{\partial x}$, which still makes sense even if $y=\phi(x)$.
In the case of $f(x,y) = e^{xy}$ we have,
$$
\frac{\partial f(x,y)}{\partial x} = ye^{xy}
$$
and
\begin{align}
\frac{df(x,y)}{dx} &= ye^{xy} + xe^{xy}\phi'(x)\\
&= e^{xy}\big{(}y + x\phi'(x)\big{)}
\end{align}
which is, as you have found and we would hope, equivalent to thinking of $f$ as a function of one variable $x$ to find by "product rule",
$$
\frac{df(x)}{dx} = e^{x\phi(x)}\big{(}\phi(x) + x\phi'(x)\big{)}
$$
Best Answer
What I think you mean is (for example), something like this $$\frac{\partial}{\partial x}\frac{\partial}{\partial y}w$$
This is usually denoted by $$\frac{\partial^2 w}{\partial x\partial y}$$and is defined by $$\lim_{\delta x\to 0}\lim_{\delta y \to 0}\left(\frac{f(x+\delta x,y+\delta y,z)-f(x+\delta x,y,z)-f(x,y+\delta y,z)+f(x,y,z)}{\delta x\delta y}\right)$$
As an example, if we let $f(x,y,z)=x^2y^3$, then $$\frac{\partial^2 f}{\partial x\partial y}=\frac{\partial }{\partial x}\frac{\partial }{\partial y}(x^2y^3)=\frac{\partial }{\partial x}(3x^2y^2)=6xy^2$$