(6) Does the function \[ f(x, y)=\left\{\begin{array}{ll}\frac{x y}{\sqrt{x^{2}+y^{2}}} & \text { if }(x, y) \neq(0,0) \\ 0 & \text { if }(x, y)=(0,0) \\ \text { have partial derivatives on } \mathbb{R}^{2} \text { ? } \\ \text { (7) Consider the equation } x^{4}+x^{3} y^{2}-y+y^{2}+y^{3}=1 \text {. Use implicit differentiation } \\ \text { to find }\left.\frac{d y}{d x}\right|_{(-1,1)}\end{array}\right. \]

**(6) Partial Derivatives of \( f(x,y) \) on \(\mathbb{R}^2\)** The function is given by \[ f(x,y)= \begin{cases} \frac{xy}{\sqrt{x^2+y^2}}, & (x,y)\neq (0,0), \\[1mm] 0, & (x,y)=(0,0). \end{cases} \] For \((x,y) \neq (0,0)\), the function is composed of elementary functions (polynomials and the square root) that are differentiable where the denominator \(\sqrt{x^2+y^2}\) is nonzero. At the origin \((0,0)\), we check the existence of the partial derivatives using their definition. * To compute the partial derivative with respect to \(x\) at \((0,0)\), \[ f_x(0,0)=\lim_{h\to0} \frac{f(h,0)-f(0,0)}{h}. \] For \(h\neq0\), \[ f(h,0)=\frac{h\cdot 0}{\sqrt{h^2+0}}=0. \] Thus, \[ f_x(0,0)=\lim_{h\to0} \frac{0-0}{h}=0. \] * Similarly, the partial derivative with respect to \(y\) at \((0,0)\) is \[ f_y(0,0)=\lim_{h\to0} \frac{f(0,h)-f(0,0)}{h}. \] For \(h\neq0\), \[ f(0,h)=\frac{0\cdot h}{\sqrt{0+h^2}}=0. \] Thus, \[ f_y(0,0)=\lim_{h\to0} \frac{0-0}{h}=0. \] Therefore, \( f \) has partial derivatives at every point in \(\mathbb{R}^2\), including the origin. --- **(7) Implicit Differentiation** We are given the equation \[ x^4 + x^3y^2 - y + y^2 + y^3 = 1. \] Differentiate both sides with respect to \(x\), remembering that \(y\) is a function of \(x\): 1. Differentiate \( x^4 \): \[ \frac{d}{dx}(x^4)= 4x^3. \] 2. Differentiate \( x^3 y^2 \) using the product rule: \[ \frac{d}{dx}(x^3y^2)= 3x^2y^2 + x^3\cdot 2y\,y' = 3x^2y^2 + 2x^3y\,y'. \] 3. Differentiate \(-y\): \[ \frac{d}{dx}(-y)= -y'. \] 4. Differentiate \( y^2 \) using the chain rule: \[ \frac{d}{dx}(y^2)= 2y\,y'. \] 5. Differentiate \( y^3 \): \[ \frac{d}{dx}(y^3)= 3y^2\,y'. \] Differentiating the right–hand side (which is constant \( 1 \)): \[ \frac{d}{dx}(1)= 0. \] Now, putting it all together, we have: \[ 4x^3 + 3x^2y^2 + 2x^3y\,y' - y' + 2y\,y' + 3y^2\,y' = 0. \] Collect the terms with \(y'\): \[ 4x^3 + 3x^2y^2 + y'\left(2x^3y - 1 + 2y + 3y^2 \right)= 0. \] Isolate \(y'\): \[ y'\left(2x^3y - 1 + 2y + 3y^2 \right) = -\left(4x^3 + 3x^2y^2\right), \] \[ y' = -\frac{4x^3+3x^2y^2}{2x^3y - 1 + 2y + 3y^2}. \] We now evaluate at \((x,y)=(-1,1)\). First, compute the numerator: - \(x^3 = (-1)^3 = -1\) - Thus, \(4x^3 = 4(-1)= -4\). - Also, \(x^2 = (-1)^2= 1\) and \(y^2= 1\), so \(3x^2y^2 = 3(1)(1)= 3\). So, \[ 4x^3+3x^2y^2 = -4+3 = -1. \] Taking the negative, \[ -\left(4x^3+3x^2y^2\right) = -(-1) = 1. \] Next, compute the denominator: \[ 2x^3y - 1 + 2y + 3y^2. \] We already have \(x^3= -1\) and \(y=1\), so: \[ 2(-1)(1)= -2. \] Also, \(2y = 2\) and \(3y^2 = 3\). Therefore, \[ -2 - 1 + 2 + 3 = 2. \] Thus, \[ y' = \frac{1}{2}. \] --- **Final Answers:** - The function \( f(x,y) \) has partial derivatives at every point in \(\mathbb{R}^{2}\). - At \((-1, 1)\), the derivative \(\frac{dy}{dx}\) is \(\frac{1}{2}\).