I want to check if my work is correct.
Find the equation of the tangent line to the curve at (2, 1)$$^2^2 + 5 = 14(1) $$ solution:
The tangent is a straight line so it will be of the form: $y=ax+b(2)$
Where $a$ is the slope of the equation of the tangent to the curve $(1)$ at point $(2,1)$, and line $y=ax+b$ passes through $(2,1)$. So that, we can get a by finding the $1st$ $\frac{\mathrm{d}y}{\mathrm{d}x}$ derivative as this is the gradient of the line. Applying Implicit Differentiation for $(1)$: \begin{alignat*}{1} 2xy^2+x^2y\frac{\mathrm{d}y}{\mathrm{d}x}+5y+5\frac{\mathrm{d}y}{\mathrm{d}x}=0 \end{alignat*} So, at $(2,1)$ it reduces to: \begin{alignat*}{1} &\ 9+9\frac{\mathrm{d}y}{\mathrm{d}x}=0\\ \Leftrightarrow &\ \frac{\mathrm{d}y}{\mathrm{d}x}=-1\\ \Leftrightarrow &\ a=-1 \end{alignat*} $(2)$ become: $y=-x+b$. $(2)$ passes through $(2,1)$: \begin{alignat*}{1} &\ 1=-2+b\\ \Leftrightarrow &\ b=3 \end{alignat*} Thus, the line $y=-x+3$ is a equation of the tangent line to the curve $x^2y^2+5xy=14$ at $(2,1)$ Sol2: \begin{align} ^2^2 + 5 = 14 \end{align}
$$2xy^2 + 2x^2yy' + 5y + 5xy'=0$$ $$y'=-\frac{2xy^2+5y}{2x^2y+5x}$$ $$m=y'(2;1)=-\frac{1}{2}$$ and the equation of tangent line at the point $(2;1)$ is $y=-\frac{1}{2}x+2$