I am given that $f$ is a function that satisfies $|f(x,y)| \leq x^2 + y^2$ for all $(x,y) \in \Bbb{R^2}$ and I am trying to show that $f$ is differentiable at $(0,0)$ I have some ideas on how to proceed but Im not sure if they are correct
to check for differentiability by definition I need to show that $lim_{(x,y) \to(0,0)}\frac{|R_{1,(a,b)}(x,y)|}{\sqrt{x^2+y^2}}=0$ well then $lim_{(x,y) \to(0,0)}\frac{|R_{1,(a,b)}(x,y)|}{\sqrt{x^2+y^2}}=\frac{|f(x,y)-f(0,0)-f_x(0,0)x-f_y(0,0)y|}{\sqrt{x^2+y^2}} \leq \frac{|f(x,y)|-|f(0,0)|-|f_x(0,0)x|-|f_y(0,0)y|}{\sqrt{x^2+y^2}}$ now my question is can I claim that $|f_x(x,y)| \leq 2x$ and similarly that $|f_y(x,y)| \leq 2y$
You are almost there. You need a linear map $L:\mathbb{R}^2\rightarrow\mathbb{R}$ such that $$\lim_{\vert\vert(x,y)\vert\vert\rightarrow0}\frac{\vert f(x,y)-f(0,0)-L(x,y)\vert}{\vert\vert(x,y)\vert\vert}\rightarrow0.$$ In this case your guess is that the $L$ is identically $0$ (because the partial derivatives in $0$ should be both $0$). Moreover the inequality gives also $f(0,0)=0$, thus $$\lim_{\vert\vert(x,y)\vert\vert\rightarrow0}\frac{\vert f(x,y)-f(0,0)-L(x,y)\vert}{\vert\vert(x,y)\vert\vert}=\lim_{\vert\vert(x,y)\vert\vert\rightarrow0}\frac{\vert f(x,y)\vert}{\vert\vert(x,y)\vert\vert}\leq \lim_{\vert\vert(x,y)\vert\vert\rightarrow0}\frac{\vert\vert(x,y)\vert\vert^2}{\vert\vert(x,y)\vert\vert}=0$$
Edit
As I understand it from the comments you want to prove that $\lim_{(x,y) \to(0,0)}\frac{|R_{1,(a,b)}(x,y)|}{\sqrt{x^2+y^2}}=0,$ where $R_{1,(a,b)}(x,y)=f(x,y)-f(0,0)-f_x(0,0)-f_y(0,0)=f(x,y)-f_x(0,0)-f_y(0,0)$. To do this we have to prove that the partial derivatives of $f$ exist at $0$. Let's see that $f_x(0,0)$ exists (and it is $0$). $$\lim_{h \rightarrow 0}\frac{\vert f(h,0)-f(0,0)\vert}{h}\leq\lim_{h \rightarrow 0}\frac{h^2}{h}=0.$$ Similarly for $f_y(0,0)$. Thus $R_{1,(a,b)}(x,y)=f(x,y)$ and you can argue as the first part of my answer to compute the limit.