Hallar puntos criticos de la función. \( f(x, y)=y^{3}-3 x^{2} y-5 x^{2}-5 y^{2}+2 \)

Find the first order derivative with respect to \( x \) for \( y^3-3x^2y-5x^2-5y^2+2 \). Evaluate the derivative by following steps: - step0: Evaluate the derivative: \(\frac{d}{dx}\left(y^{3}-3x^{2}y-5x^{2}-5y^{2}+2\right)\) - step1: Calculate: \(-6xy-10x\) Find the first order derivative with respect to \( y \) for \( y^3-3x^2y-5x^2-5y^2+2 \). Evaluate the derivative by following steps: - step0: Evaluate the derivative: \(\frac{d}{dy}\left(y^{3}-3x^{2}y-5x^{2}-5y^{2}+2\right)\) - step1: Calculate: \(3y^{2}-3x^{2}-10y\) Para encontrar los puntos críticos de la función \( f(x, y) = y^{3} - 3x^{2}y - 5x^{2} - 5y^{2} + 2 \), primero necesitamos encontrar las derivadas parciales de la función con respecto a \( x \) y \( y \). La derivada parcial de \( f \) con respecto a \( x \) es: \[ \frac{\partial f}{\partial x} = -6xy - 10x \] La derivada parcial de \( f \) con respecto a \( y \) es: \[ \frac{\partial f}{\partial y} = 3y^{2} - 3x^{2} - 10y \] Para encontrar los puntos críticos, necesitamos resolver el sistema de ecuaciones formado por estas dos derivadas parciales igualadas a cero. Vamos a resolver este sistema de ecuaciones. Solve the system of equations by following steps: - step0: Solve using the substitution method: \(\left\{ \begin{array}{l}-6xy-10x=0\\3y^{2}-3x^{2}-10y=0\end{array}\right.\) - step1: Solve the equation: \(\left\{ \begin{array}{l}x=0\cup y=-\frac{5}{3}\\3y^{2}-3x^{2}-10y=0\end{array}\right.\) - step2: Evaluate: \(\left\{ \begin{array}{l}x=0\\3y^{2}-3x^{2}-10y=0\end{array}\right.\cup \left\{ \begin{array}{l}y=-\frac{5}{3}\\3y^{2}-3x^{2}-10y=0\end{array}\right.\) - step3: Calculate: \(\left\{ \begin{array}{l}x=0\\y=0\end{array}\right.\cup \left\{ \begin{array}{l}x=0\\y=\frac{10}{3}\end{array}\right.\cup \left\{ \begin{array}{l}x=-\frac{5\sqrt{3}}{3}\\y=-\frac{5}{3}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{5\sqrt{3}}{3}\\y=-\frac{5}{3}\end{array}\right.\) - step4: Rearrange the terms: \(\left\{ \begin{array}{l}x=-\frac{5\sqrt{3}}{3}\\y=-\frac{5}{3}\end{array}\right.\cup \left\{ \begin{array}{l}x=0\\y=0\end{array}\right.\cup \left\{ \begin{array}{l}x=0\\y=\frac{10}{3}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{5\sqrt{3}}{3}\\y=-\frac{5}{3}\end{array}\right.\) - step5: Check the solution: \(\left\{ \begin{array}{l}x=-\frac{5\sqrt{3}}{3}\\y=-\frac{5}{3}\end{array}\right.\cup \left\{ \begin{array}{l}x=0\\y=0\end{array}\right.\cup \left\{ \begin{array}{l}x=0\\y=\frac{10}{3}\end{array}\right.\cup \left\{ \begin{array}{l}x=\frac{5\sqrt{3}}{3}\\y=-\frac{5}{3}\end{array}\right.\) - step6: Rewrite: \(\left(x,y\right) = \left(-\frac{5\sqrt{3}}{3},-\frac{5}{3}\right)\cup \left(x,y\right) = \left(0,0\right)\cup \left(x,y\right) = \left(0,\frac{10}{3}\right)\cup \left(x,y\right) = \left(\frac{5\sqrt{3}}{3},-\frac{5}{3}\right)\) Los puntos críticos de la función \( f(x, y) = y^{3} - 3x^{2}y - 5x^{2} - 5y^{2} + 2 \) son: 1. \((- \frac{5\sqrt{3}}{3}, -\frac{5}{3})\) 2. \((0, 0)\) 3. \((0, \frac{10}{3})\) 4. \((\frac{5\sqrt{3}}{3}, -\frac{5}{3})\)