Let \(f(x, y) = \frac{x^2 y}{x^2 + y^2}\) for \((x, y) \neq (0, 0)\) and let \(f(0, 0) = 0\).
Prove that \(f\) is continuous at \((0, 0)\).
Prove that \(f\) is not differentiable at \((0, 0)\).
Solution
1. For every \((x,y)\ne(0,0)\),\[ |f(x,y)|\le |y|. \]
If \(x=0\), then \(f(x,y)=0\). If \(x\ne0\), then \[ |f(x,y)| =\frac{x^2|y|}{x^2+y^2} \le \frac{x^2|y|}{x^2} =|y|. \]
2. The function \(f\) is continuous at \((0,0)\).
By 1, \[ |f(x,y)-f(0,0)|=|f(x,y)|\le |y|\le \sqrt{x^2+y^2}. \] The right-hand side tends to \(0\) as \((x,y)\to(0,0)\), so \(f(x,y)\to0=f(0,0)\).
3. Both partial derivatives of \(f\) at \((0,0)\) are \(0\).
Along the coordinate axes, \(f(h,0)=0\) and \(f(0,h)=0\). Hence \[ \frac{f(h,0)-f(0,0)}h=0, \qquad \frac{f(0,h)-f(0,0)}h=0 \] for \(h\ne0\), so \(\partial_xf(0,0)=\partial_yf(0,0)=0\).
4. If \(f\) were differentiable at \((0,0)\), its derivative there would be the zero linear map.
For a differentiable function, the derivative applied to the coordinate basis vectors equals the corresponding partial derivatives. By 3, both are \(0\).
5. Along the line \(y=x\), the differentiability error ratio does not tend to \(0\).
For \(x\ne0\), \[ f(x,x)=\frac{x^3}{2x^2}=\frac x2. \] Using 4, the required remainder for differentiability would be \(f(x,x)\), and \[ \frac{|f(x,x)|}{\|(x,x)\|} =\frac{|x|/2}{\sqrt2|x|} =\frac1{2\sqrt2}. \] This is constant and nonzero.
6. Therefore \(f\) is not differentiable at \((0,0)\).
Differentiability with derivative \(0\) would require the ratio in 5 to tend to \(0\) as \((x,x)\to(0,0)\), which it does not.