There is at least one caveat to your problem that needs to be stated. At least one $a_{ij}$ such that $i+j=k$ must be non-zero. Otherwise your polynomial is of degree less than $k$. And in order for the partial derivative w.r.t. $x$ to be a polynomial of degree $k-1$ one of the $a_{ij}$ must have $i\ge 1$. Similarly for $y$.
Otherwise, show that if $i+j=k$, the partial derivative of $a_{ij}x^iy^j$ for both $x$ and $y$ is of total degree $i+j-1=k-1$.
I don't think you need to use induction here.
To actually perform the partial derivatives, note that:
$$\frac{\partial f}{\partial x} \sum_{i+j \le k} a_{ij}x^iy^j = \sum_{i+j \le k} i*a_{ij}x^{i-1}y^j$$ and vice versa for the partial derivative with respect to y. Just note that in cases where there is no $x$ the partial derivative with respect to $x$ is 0.
This is taught very poorly in calculus courses, and you're confused because the notation is sloppy.
The insight you seek is the following1:
Before I go on, it's critical that you understand the following terminology:
A ("formal") parameter is a property of the function description itself.
For example, the $a$ and $b$ in the function definition $f(a,b) = a + b$ are parameters.
An argument, or "actual" parameter, is a property of an expression that is a call to a function.
For example, the $x$ and $y$ in the expression $g(x) + g(y)$ are arguments to $g$.
Now here's the kicker: if $h(x) = x^2$ then partial and total derivatives can be different:
\begin{align*}
\frac{\partial}{\partial x} f(x, h)\ =\ 1\ \color{red}{\neq}\ 2x+1\ =\ \frac{d}{dx}f(x,h)
\end{align*}
Makes sense? :-)
I hope it doesn't, because it was sloppy.
The notation above is extremely common, but not really correct.3
Remember I just told you partial derivatives are with respect to parameters whereas total derivatives are with respect to variables. This means that, if we've defined $$f(a,b) = a + b$$ as above, then it's actually incorrect (although very common) to write $$\frac{\partial}{\partial x}f(x,h)$$ for three reasons:
$x$ is not a parameter to $f$, but an argument to it. The parameter is $a$.
The second argument to $f$ should be a number (like $h(x)$), not a function like $h$.
$f(x, h)$ is not a function, but a call to a function. It evaluates to a number.
So, to really write the above derivatives correctly, I should have written:
\begin{align*}
\left.\frac{\partial f}{\partial a}\right|_{\substack{a=x\phantom{(h)}\\b=h(x)}}\ =\ \left.1\right|_{\substack{a=x\phantom{(h)}\\b=h(x)}}\ =\ 1\ \color{red}{\neq}\ 2x + 1\ =\ \frac{d}{d x} f(x, h(x))
\end{align*}
at which point it should be obvious the two aren't the same.
Makes sense? :)
1 This should be easier to understand if you know a statically typed programming language (like C# or Java).
2 You can define partial derivatives for expressions as well, but it'd just be implicitly assuming you have a function in terms of that variable, which you are differentiating, and then evaluating at a point whose value is also denoted by that variable.
3 Notice the expression wouldn't "type-check" in a statically typed programming language.
Best Answer
It's known as finding a potential of a vector field. The vector field is $(\partial_x f,\partial_y f)$, and $f$ itself is the potential (sometimes also called the potential function).
Like here, for instance: https://mathinsight.org/conservative_vector_field_find_potential.