I don't understand what you say about derivatives and I will assume that you want the following result.
Let $f: \mathbb{R}^2 \rightarrow \mathbb{R}$ be a function such that for every $a, b \in \mathbb{R}$, $x \mapsto f(x,b)$ and $y \mapsto f(a,y)$ are polynomial functions; then $f$ is a polynomial in two variables.
This is not obvious because we don't have a good representation of $f$. The idea of the proof is to show that $f$ coincides with a polynomial at sufficiently many points.
1) There exists an infinite set $I \subset \mathbb{R}$ and an integer $N$ such that for any $a,b \in I$, the polynomials $y \mapsto f(a,y)$ and $x \mapsto f(x,b)$ are of degree bounded by $N$. This follows from the fact that $\mathbb{R}$ is not countable: if $K_n$ is the set of $z \in \mathbb{R}$ such that $x\mapsto f(x,z)$ and $y \mapsto f(z,y)$ are of degree bounded by $n$, then $\cup_{n\in \mathbb{N}} K_n = \mathbb{R}$ and one of the $K_n$ must be infinite (in fact uncountable but I don't need it).
2) Let $I$ and $N$ be as in the previous point. Let $z_1,\dots,z_{N+1}$ be $N+1$ arbitrary elements in $I$. I claim that there exists a polynomial $Q$ in two variables, of degree in $x$ and $y$ at most $N$ such that $Q$ takes the same values than $f$ in all points of the form $(z_i,z_j)$, for $1 \leq i,j \leq N+1$. Indeed, this is the analog of Lagrange interpolation in two variables. This polynomial $Q$ is defined by:
$$ Q(x,y) = \sum_{i=1}^{N+1} \sum_{j=1}^{N+1} f(z_i,z_j) \prod_{i' \neq i} \prod_{j' \neq j} \frac{(x-z_{i'})(y-z_{j'})}{(z_i-z_{i'})(z_j-z_{j'})}.$$
3) I claim that $f(x,y) = Q(x,y)$ everywhere. First, $y \mapsto f(z_i,y)$ and $y \mapsto Q(z_i,y)$ are both polynomial of degree bounded by $N$, which coincide in $N+1$ points. This shows that $f = Q$ on sets of the form $z_i \times \mathbb{R}$. Now, take any $y$ in $I$. Then $x \mapsto f(x,y)$ and $x \mapsto Q(x,y)$ are polynomial of degree bounded by $N$ and they are equal for $x$ equal to one of the $z_i$. So they are equal everywhere. This shows that $f = Q$ on $\mathbb{R} \times I$. Finally, consider an arbitrary $x \in \mathbb{R}$. Then $y \mapsto f(x,y)$ and $y \mapsto Q(x,y)$ are both polynomial, equal when $y \in I$. Since $I$ is infinite, they are equal everywhere. This concludes the proof of $Q = f$.
Best Answer
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.